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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(2x=15+25\)

\(2x=40\)

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

\(x=20\)

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

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

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

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

\(2x=3\)

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

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

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

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

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

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

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

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

\(-8x=5-9\)

\(-8x=-4\)

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

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

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

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

\(3x=13\)

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

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

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

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

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

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

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

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

\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)

`#3107.\text {DN01012007}`

\(\left(x-5\right)\cdot\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0+5\\x=3-0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy, \(x\in\left\{3;5\right\}\)

_______

\(\left(2x-8\right)\cdot\left(5-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-8=0\\5-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\div2\\x=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Vậy, \(x\in\left\{4;5\right\}\)

_______

\(7x\left(2x-14\right)=0\\ \Rightarrow\left[{}\begin{matrix}7x=0\\2x-14=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x=14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=14\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

Vậy, \(x\in\left\{0;7\right\}\)

______

\(\left(2x-4\right)\cdot\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-4=0\\6-2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\div2\\x=6\div2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy, \(x\in\left\{2;3\right\}.\)

chúc bạn học giỏi

a: \(\Leftrightarrow x^2+6x+9+x^2-4-2x-2=7\)

\(\Leftrightarrow2x^2+4x-4=0\)

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

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

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

hay \(x\in\left\{-\sqrt{3}-1;\sqrt{3}-1\right\}\)

b: \(\Leftrightarrow2x^2-x-\left(2x^2+3x-4x-6\right)=0\)

\(\Leftrightarrow2x^2-x-2x^2+x+6=0\)

=>6=0(vô lý)

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

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

=>x=-2 hoặc x=2

đ: \(\Rightarrow2x^2-2x-5x+5=0\)

=>(x-1)(2x-5)=0

=>x=1 hoặc x=5/2

(2\(x\) - 1).(2\(x\) - 5) < 0

Lập bảng ta có:

Theo bảng trên ta có: \(\dfrac{1}{2}\) < \(x\) < \(\dfrac{5}{2}\)

(3 - 2\(x\)).(\(x\) + 2) > 0

Lập bảng ta có:

Theo bảng trên ta có: - 2 < \(x\) < \(\dfrac{3}{2}\) 

(3\(x\) + 1).(5 - 2\(x\)) > 0

Lập bảng ta có:

Theo bảng trên ta có: - \(\dfrac{1}{3}\) < \(x\) < \(\dfrac{5}{2}\) 

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)

1) (2x-1)(x+3)(2-x)=0

=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0

=>x=1/2 hoặc x=-3 hoặc x=2

2)x^3 + x^2 + x + 1 = 0

=>.x^2(x+1)+(x+1)=0

=>(x^2+1)(x+1)=0

=>x^2+1=0 hoặc x+1=0 

=>                      x =-1

3) 2x(x-3)+5(x-3) =0    

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

=>2x+5=0 hoặc x-3=0

=>x=-5/2 hoặc x=3

4)x(2x-7)-(4x-14)=0

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

=> x-2 =0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

5)2x^3+3x^2+2x+3=0

=>x^2(2x+3)+2x+3=0

=>(x^2+1)(2x+3)=0

=>x^2+1=0 hoặc 2x+3=0

=>                      x =-3/2

x = 3/2 đó mình chắc chắn 100 %

X= 3/2 nha bạn, chắc chắn đó.

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

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

a: (2x+1)(3-x)(4-2x)=0

=>(2x+1)(x-3)(x-2)=0

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

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

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

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

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

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

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

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

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

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

GIÚP TỚ VỚI 

a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)

bạn tin lúc trước tớ nói không tớ sai ở chổ 1x0 đóooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)