Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

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

chưa biết

Bạn vào biểu tượng \(\Sigma\) để nhập biểu thức cho chính xác nhé

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

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

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

a) Ta có: \(P=\left(\dfrac{x^2-1}{x^4-x^2+1}+\dfrac{2}{x^6+1}-\dfrac{1}{x^2+1}\right)\cdot\left(x^2-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)

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

\(=\dfrac{x^4-1+2-x^4+x^2-1}{\left(x^2+1\right)\cdot\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+x^4+x^2-x^4-x^2+1}{x^4+x^2+1}\)

\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+1}{x^4+x^2+1}\)

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

\(=\dfrac{x^2}{x^4+x^2+1}\)

\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)

ĐKXĐ : \(x\ne0\) 

Ta có \(pt\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)

Đặt \(x^2+\frac{1}{x^2}=a\) thay vào pt trên ta có :

\(pt\Leftrightarrow8\left(a+2\right)+4a^2-4.a.\left(a+2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8a+16+4a^2-4a^2-8a=\left(x+4\right)^2\)

\(\Leftrightarrow\left(x+4\right)^2=16\Leftrightarrow\orbr{\begin{cases}x+4=4\\x+4=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=0\left(KTMĐKXĐ\right)\\x=-8\left(TMĐKXĐ\right)\end{cases}}}\)

Vậy \(x=-8\)

\(\)

ko biet vua chia tay nen ko tra loi dc huhu em oi