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\(\text{Δ}=\left(-5\right)^2-4\cdot\left(-11\right)\cdot2=88+25=113>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{5-\sqrt{113}}{-22}=\dfrac{-5+\sqrt{113}}{22}\\x_2=\dfrac{-5-\sqrt{113}}{22}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^2+\left(2x-x\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+2x-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x-x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\\3x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow\left(4x+3\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(4x+3+2x-2\right)\left(4x+3-2x+2\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
\(\left(4x+3\right)^2=4x^2-8x+4\)
\(\Leftrightarrow\left(4x+3\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow\left(4x+3\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)\left(6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\6x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-1}{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)
⇔ 5x + 3 = 2x + 3
⇔ 3x = 0
⇔ x = 0
Vậy phương trình có nghiệm là x = 0
Mình làm lại rồi nhé!
a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)
⇔ 5x + 3 = 2x + 3
⇔ 3x = 0
⇔ x = 0
Vậy phương trình có nghiệm là x = 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
27x2(y - 1) - 9x3(1 - y)
= 27x2(y - 1) + 9x3(y - 1)
= (27x2 + 9x3)(y - 1)
= 9x2(3 + x)(y - 1)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(2x^2+1\ge1>0\forall x\)
\(\Rightarrow\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)
\(\Leftrightarrow4x-3=x-12\Leftrightarrow3x=-9\Leftrightarrow x=-3\)
vậy \(x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x+3\right)\left(x-4\right)\)