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\(20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)^2+48\left(\dfrac{x^2-4}{x^2-1}\right)=0\)
\(\Leftrightarrow20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)^2+48\left(\dfrac{x-2}{x+1}\right)\left(\dfrac{x+2}{x-1}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-2}{x+1}=a\\\dfrac{x+2}{x-1}=b\end{matrix}\right.\)thì ta có
\(20a^2-5b^2+48ab=0\)
\(\Leftrightarrow\left(10a-b\right)\left(2a+5b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=10a\\5b=2a\end{matrix}\right.\)
Rồi thế vô giải tiếp đi. Còn lại đơn giản nên tự làm nhé
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
đkxđ: x khác 0
\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)
\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)
\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)
\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)
\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)
\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)
=> x=0 ( loại , ko tm)
Vậy pt vô nghiệm
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
Phương pháp:
Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)
Thay vào pt
\(x\ne0:đặt:x+\dfrac{1}{x}=t\)
\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)
\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)
\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)
\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)
(bn xem lại đề)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
a. \(x-\dfrac{x+2}{3}< 3x+\dfrac{x}{2}+5\)
\(\Leftrightarrow\dfrac{6x}{6}-\dfrac{2\left(x+2\right)}{6}< \dfrac{18x}{6}+\dfrac{3x}{6}+\dfrac{30}{6}\)
\(\Rightarrow6x-2x-4-18x-3x-30< 0\)
\(\Leftrightarrow-17x< 34\)
\(\Leftrightarrow x>-2\)
b. \(\dfrac{x}{2}+\dfrac{1-x}{3}>0\)
\(\Leftrightarrow3x+2-2x>0\)
\(\Leftrightarrow x>-2\)
c. \(\left(x-9\right)^2-x\left(x+9\right)< 0\)
\(\Leftrightarrow x^2-18x+81-x^2-9x< 0\)
\(\Leftrightarrow-27x< -81\)
\(\Leftrightarrow x>3\)