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

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!

máy tính sẵn sàng

cậu giỏi nhỉ

\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)

Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\)

\(Pt\Leftrightarrow10a^2+b^2-11ab=0\\ \Leftrightarrow10a^2-10ab-ab+b^2=0\\ \Leftrightarrow10a\left(a-b\right)-b\left(a-b\right)=0\\ \Leftrightarrow\left(10a-b\right)\left(a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10a-b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10a=b\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{10\left(x-2\right)}{x+1}=\frac{x+2}{x-1}\left(1\right)\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow10\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow10\left(x^2-3x+2\right)=x^2+3x+2\\ \Leftrightarrow9x^2-33x+18=0\\ \Leftrightarrow9x^2-27x-6x+18=0\\ \Leftrightarrow9x\left(x-3\right)-6\left(x-3\right)=0\\ \Leftrightarrow\left(9x-6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\9x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{2}{3}\end{matrix}\right.\left(Tm\right)\)

\(\left(2\right)\Leftrightarrow\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow x^2-3x+2=x^2+3x+2=0\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\left(Tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{0;3;\frac{2}{3}\right\}\)

\(\text{b) }\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\\ DKXD:x\ne2;x\ne4\\ \Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-2}\cdot\frac{x-2}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)

Đặt \(\frac{x+1}{x-2}=a;\frac{x-2}{x-4}=b\)

\(Pt\Leftrightarrow a^2+ab-12b^2=0\\ \Leftrightarrow a^2+4ab-3ab-12b^2=0\\ \Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\\ \Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=-4b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-2}=\frac{3\left(x-2\right)}{x-4}\left(1\right)\\\frac{x+1}{x-2}=\frac{-4\left(x-2\right)}{x-4}\left(2\right)\end{matrix}\right.\)

Tự giải tiếp nha.

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn