ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\frac{x-2}{x+1}=a\\\frac{x+2}{x-1}=b\end{matrix}\right.\) pt trở thành:

\(5a^2-44b^2+12ab=0\) \(\Leftrightarrow\left(a-2b\right)\left(5a+22b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\5a=-22b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x-2}{x+1}=\frac{2x+4}{x-1}\\\frac{5x+10}{x-1}=\frac{-22x-44}{x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-2\right)-\left(2x-4\right)\left(x+1\right)=0\\\left(5x+10\right)\left(x-1\right)+\left(22x+44\right)\left(x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

máy tính sẵn sàng

cậu giỏi nhỉ

Nguyễn Huy Thắng á hả

f) ĐKXĐ: \(x\ge-\frac{3}{2}\)

Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)

Lũy thừa 6 cả 2 vế lên PT tương đương:

\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)

Cái ngoặc to vô nghiệm vì nó tương đương:

\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))

Vậy x = 3.

PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra

@Akai Haruma, @Nguyễn Việt Lâm

giúp em vs ạ! Cần gấp ạ

em cảm ơn nhiều!

a/ \(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)

Điều kiện: \(\left[\begin{matrix}x\le-2\\x>1\end{matrix}\right.\)

Xét \(x\le-2\) thì ta có

\(\left(x-1\right)\left(x+2\right)+4\left(x-1\right)\sqrt{\frac{x+2}{x-1}}=12\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-4\sqrt{\left(x-1\right)\left(x+2\right)}=12\)

Đặt \(\sqrt{\left(x-1\right)\left(x+2\right)}=a\left(a\ge0\right)\) thì pt thành

\(a^2-4a-12=0\)

\(\Leftrightarrow\left[\begin{matrix}a=-2\left(l\right)\\a=6\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(x-1\right)\left(x+2\right)}=6\)

\(\Leftrightarrow x^2+x-38=0\)

\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{2}+\frac{3\sqrt{17}}{2}\left(l\right)\\x=-\frac{1}{2}-\frac{3\sqrt{17}}{2}\end{matrix}\right.\)

Trường hợp x > 1 làm tương tự nhé

\(\left(1\right)\Leftrightarrow2x-3x^2+11-33x=6x-4-15x^2+10x\)

\(\Leftrightarrow12x^2-47x+15=0\)

\(\Delta=47^2-4.12.15=1489,\sqrt{\Delta}=\sqrt{1489}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{47+\sqrt{1489}}{24}\\x=\frac{47-\sqrt{1489}}{24}\end{cases}}\)

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

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

\(\Leftrightarrow x^2-6x+9-x^2-6x-9=-5\)

\(\Leftrightarrow-12x=-5\Leftrightarrow x=\frac{5}{12}\)

(2-3x)(x+11)=(3x-2)(2-5x)

<=>(3x-2)(2-5x)-(2-3x)(x+11)=0

<=>(3x-2)(2-5x)+(3x-2)(x+11)=0

<=>(3x-2)[2-5x+x+11]=0

<=>(3x-2)(13-4x)=0

<=>\(\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}\)

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

Đk:\(x\ne-3;x\ne3\)(*)

Với đk trên pt tương đương với:

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

\(x^2-6x+9-x^2-6x-9=-5.-12x=-5\)

\(x=\frac{15}{12}\left(tmđk\right)\)(*)

a)\(ĐKXĐ:\hept{\begin{cases}x>3\\x\le-1\end{cases}}\)
TH1: \(x-3>0\)
 \(\left(x-3\right)\left(x+1\right)+4.\frac{x-3}{\sqrt{x-3}}\sqrt{x+1}=-3\)
\(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Đặt \(t=\sqrt{\left(x-3\right)\left(x+1\right)}\left(t\ge0\right)\)
Phương trình trở thành:
\(t^2+4t+3=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=-3\end{cases}}\)(ktm)=> Vô Nghiệm
TH2: \(x-3< 0\)
\(\left(x-3\right)\left(x+1\right)-4.\frac{3-x}{\sqrt{3-x}}\sqrt{-x-1}=-3\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)-4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)
Tự làm tiếp nhé
 

b)Nhân chéo chuyển vế rút gọn ta được:
\(x^3-2x^2+3x-2=0\)
\(\Leftrightarrow x\left(x^2-2x+1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2\right)=0\)
\(\Rightarrow x=1\)

A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)

A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)

Rồi tiếp tục làm nhé bạn.

+\(1\le x\le4\)

+ \(\left(x-1\right)+\left(4-x\right)+2\sqrt{\left(x-1\right)\left(4-x\right)}=8+\frac{29}{5-x}\)

Với \(1\le x\le4\) ta có:

\(VT\le\left(1+1\right)\left(x-1+4-x\right)=6\)

\(VP>8\) vì \(\frac{29}{5-x}>0\)

=> PT vô nghiệm

AYUASGSHXHFSGDB HAGGAHAJF