\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

\(x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(x-2\right)=0\)

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

\(x\left(x-1\right)+5x-5=0\)

\(x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(x+5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)

\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

\(x=\frac{1}{2}\)

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

Cảm ơn

tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho

Đề là gì v bn?

a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

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

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

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

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

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)

Giúp mình nhé các bạn mình đang cần gấp lắm

1: \(\left(x-3\right)\left(2x-5\right)-3x\left(x+4\right)\)

\(=2x^2-5x-6x+15-3x^2-12x\)

\(=-x^2-23x+15\)

2: \(\left(\dfrac{1}{2}x+5\right)\left(2x-\dfrac{1}{5}\right)\)

\(=x^2-\dfrac{1}{10}x+10x-1\)

\(=x^2+\dfrac{99}{10}x-1\)

Thế mày làm đi

cho ít thôi thì làm

a) Ta có: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0\)

\(\Leftrightarrow2x^2+9x-5-2x^2+x+3=0\)

\(\Leftrightarrow10x-2=0\)

hay 10x=2

\(\Leftrightarrow x=\frac{1}{5}\)

Vậy: \(x=\frac{1}{5}\)

b) Ta có: \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\)

\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)

\(\Leftrightarrow2x-6=0\)

hay 2x=6

\(\Leftrightarrow x=3\)

Vậy: x=3

c) Ta có: \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow33x=1\)

hay \(x=\frac{1}{33}\)

Vậy: \(x=\frac{1}{33}\)

d) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

\(\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\)

\(\Leftrightarrow3x^2-x-10=2x^2-2x-4\)

\(\Leftrightarrow3x^2-x-10-2x^2+2x+4=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

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

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

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

Vậy: \(x\in\left\{-3;2\right\}\)

đ) Ta có: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

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

\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)

e) Ta có: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)

\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)

Vì \(3\ne0\)

nên \(\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: \(x\in\left\{0;4\right\}\)

a) $(x+5)(2x-1)=(2x-3)(x+1)$

$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$

$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$

b)

$(x+1)(x+9)=(x+3)(x+5)$

$\Leftrightarrow x^2+10x+9=x^2+8x+15$

$\Leftrightarrow 2x=6\Rightarrow x=3$

c)

$(3x+5)(2x+1)=(6x-2)(x-3)$

$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$

$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$

a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)

=>x(2x+10-x+2)=0

=>x(x+12)=0

=>x=0 hoặc x=-12

b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

=>(2x-5)(3x+12)=0

=>x=5/2 hoặc x=-4

c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)

=>(x-3)(3x+3)=0

=>x=3 hoặc x=-1

d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

=>(x+2)(-5x+3)=0

=>x=-2 hoặc x=3/5

\(a,\left(x-2\right)x=2x\left(x+5\right)\)

\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)

\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)

a) \(A\left(x\right)=x^7-2x^6+2x^3-2x^4-x^7+x^5+2x^6-x+5+2x^4-x^5\)

\(A\left(x\right)=(x^7-x^7)+(-2x^6+2x^6)+2x^3+(-2x^4+2x^4)+(x^5-x^5)-x+5\)

\(A\left(x\right)=2x^3-x+5\)

-  Bậc của đa thức A(x) là 3

 - Hệ số tự do: 5

- Hệ số cao nhất: 2

b) \(B\left(x\right)=-3x^5+4x^4-2x+\dfrac{1}{2}-2x^4+3x-x^5-2x^4+\dfrac{5}{2}+x\)

\(B\left(x\right)=(-3x^5-x^5)+(4x^4-2x^4-2x^4)+(-2x+x+3x)+\left(\dfrac{1}{2}+\dfrac{5}{2}\right)\)

\(B\left(x\right)=-4x^5+2x+3\)

-  Bậc của đa thức B(x) là 5

 - Hệ số tự do: 3

- Hệ số cao nhất: \(-4\)

c) \(C\left(y\right)=5y^2-2.\left(y+1\right)+3y.\left(y^2-2\right)+5\)

   \(C\left(y\right)=5y^2-2y-2+3y\left(y^2-2\right)+5\) 

   \(C\left(y\right)=5y^2-2y-2+3y^3-6y+5\)

   \(C\left(y\right)=5y^2-2y+3+3y^3-6y\)

   \(C\left(y\right)=5y^2-8y+3+3y^3\)

   \(C\left(y\right)=3y^3+5y^2-8y+3\)

-  Bậc của đa thức C(y) là 3

 - Hệ số tự do: 3

- Hệ số cao nhất: 3

a: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)

\(\Leftrightarrow x=25\)

b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x=1\)

hay \(x=-\dfrac{1}{9}\)

a: ta có: $\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15$

$\iff 2x^2+4x-5x-10-2x^2+2x=15$

$\iff x=25$

b: Ta có: $\left(5-2x\right)\left(2x+7\right)=4x^2-25$

$\iff 4x^2-25+($