mk mới lớp 6 thôi ,lớp 9 mình .......mình.........chịu (I VERY SORRY YOU!!)

sorry, i cant do it

\(\text{ĐK: }x^3-6x^2+12x-8=\left(x-2\right)^3\ne0\Leftrightarrow x\ne2\)

\(pt\Leftrightarrow\frac{\left(x-2\right)^3\left(x^2-3x-3\right)}{\left(x-2\right)^3}=0\Leftrightarrow x^2-3x-3=0\)

Vậy pt có 2 nghiệm \(a;b\) thỏa \(a+b=3;\text{ }a.b=-3\text{ (Vi-et)}\)

\(A=\frac{1}{a^{10}}+\frac{1}{b^{10}}=\frac{a^{10}+b^{10}}{\left(ab\right)^{10}}=\frac{\left(a^5+b^5\right)^2-2a^5b^5}{\left(-3\right)^{10}}\)

Ta có: \(a^5+b^5=\left(a+b\right)\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\)

\(=\left(a+b\right)\left[\left(a^4+b^4+2a^2b^2\right)-a^2b^2-ab\left(a^2+b^2\right)\right]\)

\(=\left(a+b\right)\left[\left(a^2+b^2\right)^2-ab\left(a^2+b^2\right)-a^2b^2\right]\)

\(=\left(a+b\right)\left\{\left[\left(a+b\right)^2-2ab\right]^2-ab\left[\left(a+b\right)^2-2ab\right]-\left(ab\right)^2\right\}\)

\(=3\left[\left(3^2-2.\left(-3\right)\right)^2-\left(-3\right)\left(3^2-2.\left(-3\right)\right)-\left(-3\right)^2\right]\)

\(=783\)

\(A=\frac{783^2-2\left(-3\right)^5}{3^{10}}=\frac{2525}{243}\)

a: ĐKXĐ: x∈R

\(\frac{5}{x^2-2x+2}-\frac{8}{x^2-2x+5}=3\)

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

=>\(3\left(x^2-2x+2\right)\left(x^2-2x+5\right)=5x^2-10x+25-8x^2+16x-16=-3x^2+6x+9\)

=>\(3\left\lbrack\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10\right\rbrack=-3\left(x^2-2x\right)+9\)

=>\(\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10=-\left(x^2-2x\right)+3\)

=>\(\left(x^2-2x\right)^2+8\left(x^2-2x\right)+7=0\)

=>\(\left(x^2-2x+1\right)\left(x^2-2x+7\right)=0\)

=>\(x^2-2x+1=0\)

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

=>x-1=0

=>x=1(nhận)

b: ĐKXĐ: x<>0

\(\frac{x^2-4x+3}{2x}+\frac{x^2+12x+3}{x^2+3}=4\)

=>\(\frac{x^2+3}{2x}-2+\frac{12x}{x^2+3}+1=4\)

=>\(\frac{x^2+3}{2x}+\frac{12x}{x^2+3}=4+2-1=6-1=5\)

=>\(\frac{\left(x^2+3\right)^2+24x^2}{2x\left(x^2+3\right)}=5\)

=>\(\left(x^2+3\right)^2+24x^2-10x\left(x^2+3\right)=0\)

=>\(\left(x^2+3\right)^2-4x\left(x^2+3\right)-6x\left(x^2+3\right)+24x^2=0\)

=>\(\left(x^2+3\right)\left(x^2+3-4x\right)-6x\left(x^2+3-4x\right)=0\)

=>\(\left(x^2-6x+3\right)\left(x^2-4x+3\right)=0\)

TH1: \(x^2-6x+3=0\)

=>\(x^2-6x+9-6=0\)

=>\(\left(x-3\right)^2=6\)

=>\(\left[\begin{array}{l}x-3=\sqrt6\\ x-3=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt6+3\left(nhận\right)\\ x=-\sqrt6+3\left(nhận\right)\end{array}\right.\)

TH2: \(x^2-4x+3=0\)

=>\(x^2-x-3x+3=0\)

=>(x-1)(x-3)=0

=>x=1(nhận) hoặc x=3(nhận)

6) ĐKXĐ: \(x\le-6\)

\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)

\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)

Vậy \(x\le-6\)

7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)

\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)

Vậy \(x\ge\dfrac{2}{3}\)

8) ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)

\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)

9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

dùng phương pháp đặt ẩn phụ

\(x^3-2\sqrt{2}x^2+6x-4\sqrt{2}=0\)

\(\Leftrightarrow\left(x^3-\sqrt{2}x^2+4x\right)-\left(\sqrt{2}x^2+2x-4\sqrt{2}\right)=0\)

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

\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x^2-\sqrt{2}x+4\right)=0\)

\(\Leftrightarrow x=\sqrt{2}\)

câu d tách hđt r đánh giá . VP=(x-6)^2+2>=2 còn VP <=2 =>....
câu c tương tự 
câu b c bình phương oặc đặt ẩn :3