Đầy tiên ta đi rút gọn biểu thức.

Có : $A = (3x+5).(2x-1) + (4x-1).(3x+2)$

$ = 6x^2 + 7x - 5 + 12x^2 + 5x - 2$

$ = 18x^2 + 12x-7$

Vì $|x| = 2$ nên $x = 2$ hoặc $x=-2$

Với $x=2$ ta có : $A = 18.2^2 + 12.2-7 = 89$

Với $x=-2$ ta có : $A = 18.(-2)^2 + 12.(-2) - 7 = 41$

Cảm ơn bạn nha <3

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)

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

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

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

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

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

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

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

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) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

ĐKXĐ: x<>a; x<>-a

\(\frac{ax+a}{a-x}-\frac{x-a}{a+x}=\frac{a\left(3x+1\right)}{a^2-x^2}\)

=>\(\frac{\left(ax+a\right)\left(a+x\right)+\left(a-x\right)^2}{\left(a-x\right)\left(a+x\right)}=\frac{a\left(3x+1\right)}{\left(a-x\right)\left(a+x\right)}\)

=>\(\left(ax+a\right)\left(a+x\right)+\left(a-x\right)^2=a\left(3x+1\right)\)

=>\(a^2\cdot x+x^2\cdot a+a^2+a\cdot x+a^2-2\cdot a\cdot x+x^2=3xa+a\)

=>\(x^2\left(a+1\right)+x\left(a^2-a\right)+2a^2=3xa+a\)

=>\(x^2\left(a+1\right)+x\left(a^2-4a\right)+2a^2-a=0\) (1)

TH1: a=-1

(1) sẽ trở thành:

\(x^2\left(-1+1\right)+x\cdot\left\lbrack\left(-1\right)^2-4\cdot\left(-1\right)\right\rbrack+2\cdot\left(-1\right)^2-\left(-1\right)\) =0

=>x*[1+4]+2+1=0

=>5x+3=0

=>x=-3/5(nhận)

TH2: a<>-1

\(\Delta=\left(a^2-4a\right)^2-4\cdot\left(a+1\right)\left(2a^2-a\right)\)

\(=a^4-8a^3+16a^2-4\left(2a^3-a^2+2a^2-a\right)\)

\(=a^4-8a^3+16a^2-8a^3-4a^2+4a=a^4-16a^3+12a^2+4a\)

Nếu Δ>0 thì phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-a(a-4)+\sqrt{a(a^3 - 16a^2 + 12a + 4)}}{2(a+1)}\\ x=\frac{-a(a-4)-\sqrt{a(a^3 - 16a^2 + 12a + 4)}}{2(a+1)}\end{array}\right.\)

nghiệm hay j

Tính j vậy bạn

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`

tính theo kiểu hằng đẳng thức đáng nhớ ý bạn :'(

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

3x²+3x-2x²-4x=1+x²

3x²+3x-2x²-4x-x²=1

x=-1

vậy............