\(-2\left(2x-7\right)^2=2\)

\(\Rightarrow\left(2x-7\right)^2=-4\)

Mà: \(\left(2x-7\right)^2\ge0\)

=> Ko có giá trị x cần tìm

Bạn viết lại đề bài đi!  chỗ (2x-+7) có sai không vậy ?

-2(2x+7)^2=2

Đề này mới đúng nhé, xin lỗi nhiều ạ.

a) \(29.31=\left(30-1\right)\left(30+1\right)=30^2-1=900-1=899\)

b) \(33,3^2-2.33,3.3,3+3,3^2=\left(33,3-3,3\right)^2=30^2=900\)

c) \(20,1.19,9=\left(20+0,1\right)\left(20-0,1\right)=20^2-0,1^2=400-0.01=399,99\)

d) \(37^2+2.37.13+13^2=\left(37+13\right)^2=50^2=2500\)

a, \(29.31=\left(30-1\right)\left(30+1\right)=30^2-1=899\)

b, \(33,3^2-2.33,3.3,3+3,3^2=\left(33,3-3,3\right)^2=30^2=900\)

c, \(20,1.19,9=\left(20-0,1\right)\left(20+0,1\right)=20^2-0,1^2=399,99\)

d, \(37^2+2.37.13+13^2=\left(37+13\right)^2=50^2=2500\)

a: \(29\cdot31=30^2-1=900-1=899\)

b: \(33.3^2-2\cdot33.3\cdot3.3+3.3^2\)

\(=\left(33.3-3.3\right)^2\)

=900

c: \(20.1\cdot19.9=20^2-0.1^2=399.9\)

d: \(37^2+2\cdot37\cdot13+13^2=\left(37+13\right)^2=2500\)

a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.

b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.

Lời giải:

a.

$|4x-1|-|3x-\frac{1}{2}|=0$

$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)

b. Nếu $x\geq 1$ thì:

$|x-1|-2x=\frac{1}{2}$

$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$

$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)

Nếu $x< 1$ thì:

$1-x-2x=\frac{1}{2}$

$\Leftrightarrow x=\frac{1}{6}$ (tm)

Mk xin phép ko vt lại đề nx

\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)

\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)

\(\Rightarrow x=\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{3}{2}-2-\left(1-5\right)\left(\dfrac{1}{2}-1\right)=-\dfrac{5}{2}\)

1,2,3,4 không tính được.

`5)(2x-1/2)^2`

`=(2x)^2-2+(1/2)^2`

`=4x^2-2+1/4`

`6)(x+1/4)^2`

`=x^2+1/2x+1/16`

3) \(3x-2y^2=\left(\sqrt{3x}\right)^2-\left(y\sqrt{2}\right)^2=\left(\sqrt{3x}-y\sqrt{2}\right)\left(\sqrt{3x}+y\sqrt{2}\right)\)

4) \(5x-y^2=\left(\sqrt{5x}-y\right)\left(\sqrt{5x}+y\right)\)

( 1/2 + 1/3 ) x 2/5

c1 = 5/6 x 2/5 = 1/3

c2 = 1/2 x 2/5 + 1/3 x 2/5

= 1/5 + 2/15 

= 1/3

3/5 x 17/21 x 2/5

c1 := 17/35 x 2/5 = 34/175

c2 : = (3/5 x 2/5) x 17/21 

= 6/25 x 17/21 

= 34/175?

( 1/3 - 1/5 ) x 1/2

c1 : = 2/15 x 1/2 

= 1/15

c2 : = 1/3 x 1/2 - 1/5 x 1/2

= 1/6 - 1/10

= 1/15

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

Bài 1: 

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

Bài 2: 

a: \(4x^2+4x+1\)

b: \(9x^2+9x+\dfrac{9}{4}\)