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Answer 1

$\sqrt{4 x - 20} + 3 \sqrt{x - \frac{5}{9}} - \frac{1}{3} . \sqrt{9 x - 45} = 4 2 \sqrt{x - 5} + 3 . \sqrt{\frac{9 x - 5}{9}} - \frac{1}{3} . \sqrt{9 (x - 5)} = 4 2 \sqrt{x - 5} + 3 . \frac{\sqrt{9 x - 5}}{3} - \frac{3 . \sqrt{x - 5}}{3} = 4 2 \sqrt{x - 5} + \sqrt{9 x - 5} - \sqrt{x - 5} = 4 \sqrt{x - 5} + \sqrt{9 x - 5} = 4 x - 5 + 2 . \sqrt{x - 5} . \sqrt{9 x - 5} + 9 x - 5 = 16 2 . \sqrt{x - 5} . \sqrt{9 x - 5} + 10 x - 26 = 0 \sqrt{(x - 5) . (9 x - 5)} + 5 x - 13 = 0 \sqrt{(x - 5) . (9 x - 5)} = 13 - 5 x (9 x^{2} - 5 x - 45 x + 25) = 169 - 130 x + 25 x^{2} 169 - 130 x + 25 x^{2} - 9 x^{2} + 50 x - 25 = 0 16 x^{2} - 80 x + 144 = 0 {(4 x)}^{2} - 2.4 x . 10 + 100 + 14 = 0 {(4 x - 10)}^{2} + 14 = 0 (v ô l ý) v ì (4 x + 10) \geq 0 n ê n {(4 x - 10)}^{2} + 14 > 0 P h ư ơ n g t r ì n h v ô n g h ệ m$

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Answer 2

$\sqrt{4 x - 20} + 3 \sqrt{x - \frac{5}{9}} - \frac{1}{3} . \sqrt{9 x - 45} = 4 2 \sqrt{x - 5} + 3 . \sqrt{\frac{9 x - 5}{9}} - \frac{1}{3} . \sqrt{9 (x - 5)} = 4 2 \sqrt{x - 5} + 3 . \frac{\sqrt{9 x - 5}}{3} - \frac{3 . \sqrt{x - 5}}{3} = 4 2 \sqrt{x - 5} + \sqrt{9 x - 5} - \sqrt{x - 5} = 4 \sqrt{x - 5} + \sqrt{9 x - 5} = 4 x - 5 + 2 . \sqrt{x - 5} . \sqrt{9 x - 5} + 9 x - 5 = 16 2 . \sqrt{x - 5} . \sqrt{9 x - 5} + 10 x - 26 = 0 \sqrt{(x - 5) . (9 x - 5)} + 5 x - 13 = 0 \sqrt{(x - 5) . (9 x - 5)} = 13 - 5 x (9 x^{2} - 5 x - 45 x + 25) = 169 - 130 x + 25 x^{2} 169 - 130 x + 25 x^{2} - 9 x^{2} + 50 x - 25 = 0 16 x^{2} - 80 x + 144 = 0 {(4 x)}^{2} - 2.4 x . 10 + 100 + 14 = 0 {(4 x - 10)}^{2} + 14 = 0 (v ô l ý) v ì (4 x + 10) \geq 0 n ê n {(4 x - 10)}^{2} + 14 > 0 P h ư ơ n g t r ì n h v ô n g h ệ m$

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Answer 3

Phỏng đoán 1: Rút gọn biểu thức
Giả sử biểu thức là: A=4x−20+3x−59−139x−45cap A equals the square root of 4 x minus 20 end-root plus 3 the square root of the fraction with numerator x minus 5 and denominator 9 end-fraction end-root minus one-third the square root of 9 x minus 45 end-root
𝐴=4𝑥−20√+3𝑥−59−139𝑥−45√

Điều kiện xác định: x−5≥0⟹x≥5x minus 5 is greater than or equal to 0 ⟹ x is greater than or equal to 5
𝑥−5≥0⟹𝑥≥5
.
Cách rút gọn:
Phân tích các số hạng: 4x−20=4(x−5)=4x−5=2x−5the square root of 4 x minus 20 end-root equals the square root of 4 open paren x minus 5 close paren end-root equals the square root of 4 end-root the square root of x minus 5 end-root equals 2 the square root of x minus 5 end-root
4𝑥−20√=4(𝑥−5)√=4√𝑥−5√=2𝑥−5√
.
3x−59=3x−59=3x−53=x−53 the square root of the fraction with numerator x minus 5 and denominator 9 end-fraction end-root equals 3 the fraction with numerator the square root of x minus 5 end-root and denominator the square root of 9 end-root end-fraction equals 3 the fraction with numerator the square root of x minus 5 end-root and denominator 3 end-fraction equals the square root of x minus 5 end-root
3𝑥−59=3𝑥−5√9√=3𝑥−5√3=𝑥−5√
.
139x−45=139(x−5)=139x−5=13⋅3⋅x−5=x−5one-third the square root of 9 x minus 45 end-root equals one-third the square root of 9 open paren x minus 5 close paren end-root equals one-third the square root of 9 end-root the square root of x minus 5 end-root equals one-third center dot 3 center dot the square root of x minus 5 end-root equals the square root of x minus 5 end-root
139𝑥−45√=139(𝑥−5)√=139√𝑥−5√=13⋅3⋅𝑥−5√=𝑥−5√
.

Thay vào biểu thức A: A=2x−5+x−5−x−5cap A equals 2 the square root of x minus 5 end-root plus the square root of x minus 5 end-root minus the square root of x minus 5 end-root
𝐴=2𝑥−5√+𝑥−5√−𝑥−5√
.

Kết quả: A=2x−5cap A equals 2 the square root of x minus 5 end-root
𝐴=2𝑥−5√
.

Phỏng đoán 2: Giải phương trình
Giả sử biểu thức trên bằng 0: A=0cap A equals 0
𝐴=0

2x−5=0⟹x−5=0⟹x−5=0⟹x=52 the square root of x minus 5 end-root equals 0 ⟹ the square root of x minus 5 end-root equals 0 ⟹ x minus 5 equals 0 ⟹ x equals 5
2𝑥−5√=0⟹𝑥−5√=0⟹𝑥−5=0⟹𝑥=5
Nghiệm này thoả mãn điều kiện x≥5x is greater than or equal to 5
𝑥≥5
.

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Answer 4

Phỏng đoán 1: Rút gọn biểu thức
Giả sử biểu thức là: A=4x−20+3x−59−139x−45cap A equals the square root of 4 x minus 20 end-root plus 3 the square root of the fraction with numerator x minus 5 and denominator 9 end-fraction end-root minus one-third the square root of 9 x minus 45 end-root
𝐴=4𝑥−20√+3𝑥−59−139𝑥−45√

Điều kiện xác định: x−5≥0⟹x≥5x minus 5 is greater than or equal to 0 ⟹ x is greater than or equal to 5
𝑥−5≥0⟹𝑥≥5
.
Cách rút gọn:
Phân tích các số hạng: 4x−20=4(x−5)=4x−5=2x−5the square root of 4 x minus 20 end-root equals the square root of 4 open paren x minus 5 close paren end-root equals the square root of 4 end-root the square root of x minus 5 end-root equals 2 the square root of x minus 5 end-root
4𝑥−20√=4(𝑥−5)√=4√𝑥−5√=2𝑥−5√
.
3x−59=3x−59=3x−53=x−53 the square root of the fraction with numerator x minus 5 and denominator 9 end-fraction end-root equals 3 the fraction with numerator the square root of x minus 5 end-root and denominator the square root of 9 end-root end-fraction equals 3 the fraction with numerator the square root of x minus 5 end-root and denominator 3 end-fraction equals the square root of x minus 5 end-root
3𝑥−59=3𝑥−5√9√=3𝑥−5√3=𝑥−5√
.
139x−45=139(x−5)=139x−5=13⋅3⋅x−5=x−5one-third the square root of 9 x minus 45 end-root equals one-third the square root of 9 open paren x minus 5 close paren end-root equals one-third the square root of 9 end-root the square root of x minus 5 end-root equals one-third center dot 3 center dot the square root of x minus 5 end-root equals the square root of x minus 5 end-root
139𝑥−45√=139(𝑥−5)√=139√𝑥−5√=13⋅3⋅𝑥−5√=𝑥−5√
.

Thay vào biểu thức A: A=2x−5+x−5−x−5cap A equals 2 the square root of x minus 5 end-root plus the square root of x minus 5 end-root minus the square root of x minus 5 end-root
𝐴=2𝑥−5√+𝑥−5√−𝑥−5√
.

Kết quả: A=2x−5cap A equals 2 the square root of x minus 5 end-root
𝐴=2𝑥−5√
.

Phỏng đoán 2: Giải phương trình
Giả sử biểu thức trên bằng 0: A=0cap A equals 0
𝐴=0

2x−5=0⟹x−5=0⟹x−5=0⟹x=52 the square root of x minus 5 end-root equals 0 ⟹ the square root of x minus 5 end-root equals 0 ⟹ x minus 5 equals 0 ⟹ x equals 5
2𝑥−5√=0⟹𝑥−5√=0⟹𝑥−5=0⟹𝑥=5
Nghiệm này thoả mãn điều kiện x≥5x is greater than or equal to 5
𝑥≥5
.

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