\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)

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

\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

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

\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)

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

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Ta có \(\frac{9+4x^2+4x^3+x^4}{x^2+2x}=\frac{x^2\left(x^2+2\right)+2x\left(x^2+2x\right)+9}{x^2+2x}\)

= x² + 2x + \(\frac{9}{x^2+2x}\)

= (\(\frac{3}{\sqrt{x^2+2x}}-\sqrt{x^2+2x}\))² + 6 \(\ge6\)

\(\frac{9+x^2\left(x^2+2x\right)+2x\left(x^2+2x\right)}{x^2+2x}\)

Nha a viết láu táu nên thiếu mất x

em phục đến kinh hoàng, sao a2 đưa về hđt 1 cách tài tình như phu thuy,

đây la bai thi violym...lop10 cua chị em đó

\(\frac{2x^2+4x+9}{x^2+2x+4}=\frac{6x^2+12x+27}{3\left(x^2+2x+4\right)}=\frac{7\left(x^2+2x+4\right)-x^2-2x-1}{3\left(x^2+2x+4\right)}=\frac{7}{3}-\frac{\left(x+1\right)^2}{3\left(x+1\right)^2+9}\le\frac{7}{3}\)

Dấu "=" xảy ra khi \(x=-1\)

a) Ta có : \(E=2+\frac{1}{x^2+2x+4}=2+\frac{1}{\left(x+1\right)^2+3}\) đạt GTLN

\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+3}\)đạt GTLN

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

Vậy GTLN của E là \(\frac{7}{3}\)khi x = -1

\(F=\frac{6x-8}{x^2+1}=\frac{\left(x^2+1\right)-\left(x^2-6x+9\right)}{x^2+1}=1-\frac{\left(x-3\right)^2}{x^2+1}\)

F có GTLN \(\Leftrightarrow\frac{\left(x-3\right)^2}{x^2+1}\)có GTNN khi x = 3

Vậy GTLN của F là 1 khi x = 3

Cách khác cho câu b

\(F=\frac{6x-8}{x^2+1}\Rightarrow F\cdot x^2+F-6x+8=0\)

\(\Leftrightarrow F\cdot x^2-6x+\left(F+8\right)=0\)

Xét \(\Delta'=9-\left(F+8\right)\cdot F=9-F^2-8F\ge0\)

Đến đây chặn F là được nhế !!!!

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

\(A_{min}=-2\) khi \(x=-2\)

Với 2 câu B, C cần kiến thức lớp 9 để làm:

\(Bx^2+2Bx+3B=x^2-2x+2\)

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

\(\Delta'=\left(B+1\right)^2-\left(B-1\right)\left(3B-2\right)\ge0\)

\(\Leftrightarrow2B^2-7B+1\le0\Rightarrow\frac{7-\sqrt{41}}{4}\le B\le\frac{7+\sqrt{41}}{4}\)

\(B_{min}=\frac{7-\sqrt{41}}{4}\) khi \(x=\frac{\sqrt{41}-1}{4}\)

\(2Cx^2+4Cx+9C=x^2-2x-1\)

\(\Leftrightarrow\left(2C-1\right)x^2+2\left(2C+1\right)x+9C+1=0\)

\(\Delta'=\left(2C+1\right)^2-\left(2C-1\right)\left(9C+1\right)\ge0\)

\(\Leftrightarrow14C^2-11C-2\le0\Rightarrow\frac{11-\sqrt{233}}{28}\le C\le\frac{11+\sqrt{233}}{28}\)

\(C_{min}=\frac{11-\sqrt{233}}{28}\) khi \(x=\frac{\sqrt{233}-11}{8}\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)

\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)

\(< =>4x-2+x=14x+7\)

\(< =>14x-5x=-2-7\)

\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)

\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)

\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)

\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)

\(A=\frac{\frac{1}{2}\left(2x^2+4x+9\right)-\frac{11}{2}}{2x^2+4x+9}=\frac{1}{2}-\frac{11}{2}.\frac{1}{2x^2+4x+9}\)

Nhận xét: 2x² + 4x + 9 = 2.(x² + 2x + 1) + 7 = 2.(x + 1)² + 7 > 7 với mọi x

=> \(\frac{1}{2x^2+4x+9}\le\frac{1}{7}\)=> \(-\frac{11}{2}.\frac{1}{2x^2+4x+9}\ge\frac{-11}{2}.\frac{1}{7}=-\frac{11}{14}\)

=> A > \(\frac{1}{2}-\frac{11}{14}=-\frac{2}{7}\) 

Vậy A nhỏ nhất bằng -2/7 khi  x+ 1 = 0  => x = -1

bạn đưa ra là

x²+2x-1=2x²+4x+9

rồi chuyển vế là xong

​mình cũng không bik có đúng không

​mik mới học lớp 7 thôi