\(a,\Leftrightarrow x^2-x-x^2+6x+16=1\\ \Leftrightarrow5x=-15\Leftrightarrow x=-3\\ b,\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

a: \(\Leftrightarrow x-3=7\)

hay x=10

1.

a) \(2x^4-4x^3+2x^2\)

\(=2x^2\left(x^2-2x+1\right)\)

\(=2x^2\left(x-1\right)^2\)

b) \(2x^2-2xy+5x-5y\)

\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)

\(=2x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\cdot\left(2x+5\right)\)

2 . 

a,

\(4x\left(x-3\right)-x+3=0\)

⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)

⇒\(\left(x-3\right)\left(4x-1\right)=0\)

⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)

b, 

\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)

⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0

⇒\(\left(x-4\right)\left(3x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)

a) 2x² - 3x - 2 = 0.

<=> (2x + 1)(x - 2) = 0

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

Câu 1:

a: 3(x-1)-(x+1)=-1

=>3x-3-x-1=-1

=>2x-4=-1

=>2x=-1+4=3

=>\(x=\frac32\)

b: Đặt f(x)=0

=>\(2x^2-x=0\)

=>x(2x-1)=0

=>x=0 hoặc x=1/2

Bài 3:

a: Xét ΔCDA và ΔEAD có

CD=EA

\(\hat{CDA}=\hat{EAD}\) (hai góc so le trong, CD//AE)

AD chung

Do đó: ΔCDA=ΔEAD

b: Ta có: \(\hat{BAD}+\hat{CAD}=\hat{BAC}=90^0\)

\(\hat{BDA}+\hat{HAD}=90^0\) (ΔHAD vuông tại H)

mà \(\hat{CAD}=\hat{HAD}\) (AD là phân giác của góc HAC)

nên \(\hat{BAD}=\hat{BDA}\)

=>ΔBAD cân tại B

Bài 2:

a: g(x)-f(x)+h(x)

\(=-3x^3+2x^2+3x-2-2x^2+3x-x-1+2x^2+1\)

\(=-3x^3+2x^2+5x-2\)

b: f(x)=2x^2-3x-x+1=2x^2-4x+1

f(-1)=\(2\cdot\left(-1\right)^2-4\cdot\left(-1\right)+1=2+4+1=7\)

\(h\left(\frac12\right)=2\cdot\left(\frac12\right)^2+1=2\cdot\frac14+1=\frac12+1=\frac32\)

f(-1)-h(1/2)

=7-3/2

=11/2

c: f(x)=h(x)

=>\(2x^2-4x+1=2x^2+1\)

=>-4x=0

=>x=0

\(2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{2x + 1 = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = \dfrac{{ - 1}}{2}}\end{array}} \right.\)

Vậy \(x = 0;x = \dfrac{{ - 1}}{2}\)

Ta có: \(x^3-2x^2-x+2=0\)

\(\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

- Ũa được sửa đề hã ?