giup giải pt:√x-2+√4-x=2x2-5x-1
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a) \(2x^2-5x+1=0\)
\(\Delta=b^2-4ac\Rightarrow\left(-5\right)^2-4.2.1=17>0\)
Phương trình có 2 nghiệm phân biệt:
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-5\right)+\sqrt{17}}{2.2}=\dfrac{5+\sqrt{17}}{4}\)
\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-5\right)-\sqrt{17}}{2.2}=\dfrac{5-\sqrt{17}}{4}\)
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b) \(4x^2+4x+1=0\)
\(\Delta=b^2-4ac\Rightarrow4^2-4.4.1=0\)
Vậy phương trình có nghiệm kép:
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c) \(5x^2-x+2=0\)
\(\Delta=b^2-4a\Rightarrow\left(-1\right)^2-4.5.2=-39\)
Vậy phương trình vô nghiệm.
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\(x\left(3x-4\right)=2x^2+1\)
\(\Leftrightarrow3x^2-4x-2x^2-1=0\)
\(\Leftrightarrow x^2-4x-1=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=-1\end{matrix}\right.\)
Ta có :
\(A=x_1^2+x_2^2+3x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2\)
\(=\left(x_1+x_2\right)^2+x_1x_2\)
\(=4^2-1\)
\(=16-1\)
\(=15\)
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Mình phân tích ở trên rồi
Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
=> x - 1 = 0 => x = 0+1 => x = 1
Hoặc x - 2 = 0 => x = 0+2 => x = 2
Hoặc x - 3 = 0 => x = 0+3 => x = 3
Vậy x = 1 hoặc x = 2 hoặc x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2x^2+6x-4\left(x+3\right)\)
\(=\left(2x^2+6x\right)-4\left(x+3\right)\)
\(=2x\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(2x+4\right)\)
\(=2\left(x+3\right)\left(x+2\right)\)
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\(xy\left(x-y\right)-5x+5y\)
\(=xy\left(x-y\right)-\left(5x-5y\right)\)
\(=xy\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-5\right)\)
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\(2x^2+3x-4xy-6y\)
\(=\left(2x^2+3x\right)-\left(4xy+6y\right)\)
\(=x\left(2x+3\right)-2y\left(2x+3\right)\)
\(=\left(x-2y\right)\left(2x+3\right)\)
2x² + 6x - 4(x + 3)
= (2x² + 6x) - 4(x + 3)
= 2x(x + 3) - 4(x + 3)
= (x + 3)(2x - 4)
= 2(x + 3)(x - 2)
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xy(x - y) - 5x + 5y
= xy(x - y) - (5x - 5y)
= xy(x - y) - 5(x - y)
= (x - y)(xy - 5)
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2x² + 3x - 4xy - 6y
= (2x² - 4xy) + (3x - 6y)
= 2x(x - 2y) + 3(x - 2y)
= (x - 2y)(2x + 3)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)