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

Với x >= 2 thì \(\frac{x}{2}+\frac{3\left(x-2\right)}{2}=\frac{3x+x-6}{2}=\frac{4x-6}{2}=2x-3\)

Với x < 2 thì \(\frac{x}{2}+\frac{3\left(2-x\right)}{2}=\frac{x+6-3x}{2}=\frac{6-2x}{2}=3-x\)

ÁP DỤNG COSI CHO HAI SỐ KHÔNG ÂM RỒI BIỆN LUẬN SUY RA \(\left(x;y\right)=\left\{\left(1;1\right),\left(1;-1\right),\left(-1;1\right),\left(-1;-1\right)\right\}.\)

Chứng minh BĐT AM-GM cho 2 số không âm: (nếu cần): Ta cần c/m: \(a^2+b^2\ge2ab\)

Thật vậy: \(\left(a-b\right)^2\ge0\Rightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

--------------------------------------------------------------------------------

Áp dụng BĐT AM-GM cho hai số không âm:\(\left(4x^2+\frac{4}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)\ge8+2=10=VP\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}4x^2=\frac{4}{x^2}\\y^2=\frac{1}{y^2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2=\frac{1}{x^2}\\y^4=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1..\left(h\right)...x=-1\\y=1..\left(h\right)...y=-1\end{cases}}\)

Lập tiếp ra các cặp số nha!

bạn nào giúp mik vs

\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)

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

\(=>\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)

\(=>\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=-\frac{7}{6}\)

\(=>x=-\frac{7}{6}:\frac{3}{2}=-\frac{7}{9}\)

a đkxđ khi x khác 2 và -2     \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2-\left(x-2\right)^2}{x^2-4}=\frac{4}{x^2-4}\)

\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)\(\Rightarrow\left(x+2-x+2\right)\left(x+2+x-2\right)=4\Rightarrow4\cdot2x=4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)(thảo mãn)

b đkxđ khi x+3 khác 0 suy ra x khác -3

\(\frac{x^2-9}{x+3}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3=0\Rightarrow x=3\)(thảo mãn)

a, | x - 3/4 | = 1/2

=>\(\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

=>\(\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}x=\frac{2}{4}+\frac{3}{4}\\x=-\frac{2}{4}+\frac{3}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)

Vậy....

a) \(|x-\frac{3}{4}|=\frac{1}{2}\)

\(< =>\orbr{\begin{cases}x-\frac{3}{4}=\frac{1}{2}\\x-\frac{3}{4}=-\frac{1}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)

Vay : x = 5/4 hoặc x =  1/4

b)\(saide\)

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

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

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

__

`(x+2)(5-3x)=0`

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

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

__

\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

\(a)x+30\%x=-1,31\)

\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)

\(\Leftrightarrow10x+3x=-13,1\)

\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)

\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)

\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=9\)

\(\Leftrightarrow6x-3=18\)

\(\Leftrightarrow x=\frac{7}{2}\)

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)

=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)

=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)

=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)

=>x-3=0

=>x=3

Làm ngắn gọn thôi nhé :v

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

\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)

\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)

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

\(A=\frac{x+2}{x-3}\)

\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)

\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)

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

\(B=\frac{x+2}{x-2}\)

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

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

\(C=\frac{10x}{-x^2+9}\)

\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)

\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)

\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)

\(D=\frac{51x-15}{2x^3-18x}\)

\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

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

\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)

\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)

\(E=\frac{10x^2+10}{x^4-2x+1}\)