|3x-7|+|3x-2|+8 >= 5+8 = 13 

Dấu "=" xảy ra <=> 3/2 <= x <= 7/3

k mk nha

tiếp đi bạn 

Lời giải:
$A=x^2+2y^2+3x-y+6$

$\Leftrightarrow x^2+3x+(2y^2-y+6-A)=0(*)$

Coi đây là PT bậc 2 ẩn $x$

Vì $A$ xác định nên $(*)$ luôn có nghiệm.

$\Rightarrow \Delta'=9-4(2y^2-y+6-A)\geq 0$

$\Leftrightarrow A\geq 8y^2-4y+15$

Mà $8y^2-4y+15=8(y-\frac{1}{4})^2+\frac{29}{2}\geq \frac{29}{2}$

$\Rightarrow A\geq \frac{29}{2}$ hay $A_{\min}=\frac{29}{2}$
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\(B=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}\)

$x^2\geq 0\Rightarrow x^2+1\geq 1\Rightarrow \frac{2}{x^2+1}\leq 2$

$\Rightarrow B=1-\frac{2}{x^2+1}\geq 1-2=-1$

Vậy $B_{\min}=-1$

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ĐK: $x\neq 1$

\(C=\frac{x^2-3x+3}{x^2-2x+1}=\frac{x^2-2x+1-(x-1)+1}{x^2-2x+1}=1-\frac{1}{x-1}+\frac{1}{(x-1)^2}\)

\(=\left(\frac{1}{x-1}-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}\)

Vậy $C_{\min}=\frac{3}{4}$

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

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

Đặt \(\frac{1}{x-1}=c\)

\(\Rightarrow\) \(C=c^2-c+1\)

\(=c^2-2.c.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(c-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) \(\forall c\)

Vậy GTNN của C là \(\frac{3}{4}\)

Dấu '' = '' xảy ra khi \(c=\frac{1}{2}\Leftrightarrow\frac{1}{x-1}=\frac{1}{2}\Leftrightarrow3\)

a)\(ĐKXĐ:x\ne0;-1\)

Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)

d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)

e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=0\)

MIk CHỈ GIẢI A VÀ B THÔI NHÉ!! NẾU SAI MONG CÁC BẠN THÔNG CẢM!!

A= \(\left(x+y\right)^2-2xy\ge-2xy\)

B= \(3\left(x^2+y^2\right)+4xy=3\left[\left(x+y\right)^2-2xy\right]+4xy\)

  = \(3\left(x+y\right)^2-6xy+4xy=3\left(x+y\right)^2-2xy\ge-6xy\)( DO TỚ LẤY 3 NHÂN VỚI -2 NHA)

VẬY GTNN CỦA A VÀ B LẦN LƯỢT LÀ -2XY VÀ -6XY (ĐỀU TMĐK) 

) \(\dfrac{x^3+8y^3}{2y+x}\)

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

\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)

\(=x^2+2xy+4y^2\)

b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)

\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)

\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)

\(=\dfrac{3a-1}{2\left(a-4\right)}\)

c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)

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

\(=\dfrac{\left(x+y\right)^2}{2}\)

d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)

\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)

\(=x^2-10x+25+7x+14-x^2-2x\)

\(=39-5x\)

e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)

\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)

\(=\dfrac{3x+2x+1}{x-2}\)

\(=\dfrac{5x+1}{x-2}\)

h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)

\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)

\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)

\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

a)có khả năng sai đề bài

b)Liệu có sai đề bài không

c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)

\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)(Hơi tắt)

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

d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x-2y}\)