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3 tháng 7 2016

a, 2(x+5)=x2+5x

=> 2x+10=x2+5x

=> 0=x2+5x-2x-10

=> x2+3x-10=0

=> x2+5x-2x-10=0

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

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

=> x-2 =0 hoặc x+5 =0

=> x=2 hoặc x=-5

b, 4x2-25=(2x-5)(2x+7)

=> (2x)2-52=(2x-5)(2x+7)

=> (2x-5)(2x+5) - (2x-5)(2x+7)=0

=> (2x-5)(2x+5-2x-7)=0

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

=> 2x-5=0

=> 2x=5

=> x =2,5

c, x3+x=0

=>x(x2+1)=0

=> x=0 hoặc x2+1=0

Mà x2+1 >= 1 nên x=0

d, Hình như là thiếu đề

3 tháng 7 2016

a,=2x+10=x2+5x

   =-x2-2x-5x+10=0

   =-x2-7x+10=0

   Delta=(-7)2-4.-1.10=89

x1=7+căn89/2      x2=7-căn 89/2

CÁC CÂU KHÁC TỰ GIẢI NHA bạn

a. (3x - 1)2 - (x + 3)2 = 0

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

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

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

 

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

13 tháng 11 2021

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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

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

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

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)

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

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

14 tháng 2 2022

1.

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

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

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

9 tháng 12 2021

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

30 tháng 10 2021

c: \(x^2+4x+4=\left(x+2\right)^2\)

d: \(9x^2+6x+1=\left(3x+1\right)^2\)

13 tháng 11 2021

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

7 tháng 8 2021

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1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x

2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)

b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)

c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)

30 tháng 9 2018

\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

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

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

\(=4x^2\)

hk tốt

^^

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

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

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

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