Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
e:
Tham khảo: 
a: \(\Leftrightarrow x^2-2x+1+4x^2+4x+4-5x^2+5=0\)
\(\Leftrightarrow2x+10=0\)
hay x=-5
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
c) \(8x^3-1=8x^2+4x+2\)
<=> \(\left(2x-3\right)\left(4x^2+2x+1\right)=0\)
<=> \(2x-3=0\) hoặc \(4x^2+2x+1=0\)
Th1: x=\(\dfrac{3}{2}\)
Th2: Vô nghiệm
Vậy x=\(\dfrac{3}{2}\)
\(\text{a) }\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\\ \Leftrightarrow\left(\dfrac{2x^2-x-1}{2}-3x^2+x+4\right)2=\left(5-x\right)\left(2x+4\right)2\\ \Leftrightarrow2x^2-x-1-6x^2+2x+8=\left(5-x\right)\left(4x+8\right)\\ \Leftrightarrow-4x^2+x+7=20x+40-4x^2-8x\\ \Leftrightarrow-4x^2+x+4x^2-12x=40-7\\ \Leftrightarrow-11x=33\\ \Leftrightarrow x=-3\\ \text{Vậy }S=\left\{-3\right\}\)
\(\text{b) }\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\left(x-1\right)\left(x+2\right)+1\\ \Leftrightarrow\left(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1\right)4=\left(x^2-x+2x-2+1\right)4\\ \Leftrightarrow\left(2x-5\right)\left(3x+7\right)+8x-4=\left(x^2+x-1\right)4\\ \Leftrightarrow6x^2-15x+14x-35+8x-4=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-39=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-4x^2-4x-39+4=0\\ \Leftrightarrow2x^2+3x-35=0\\ \Leftrightarrow2x^2+10x-7x-35=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(7x+35\right)=0\\ \Leftrightarrow2x\left(x+5\right)-7\left(x+5\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-5\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{\dfrac{7}{2};-5\right\}\)
\(\text{c) }8x^3-1=8x^2+4x+2\\ \Leftrightarrow\left(2x-1\right)\left(4x^2+2x+1\right)=2\left(4x^2+2x+1\right)\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }S=\left\{\dfrac{3}{2}\right\}\)
\(\text{d) }\left(x^2+x+1\right)\left(x^2-x+1\right)=x^6-1\\ \Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)=1\\ \Leftrightarrow x^2-1=1\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\sqrt{2}\\ \text{Vậy }S=\left\{\sqrt{2}\right\}\)
\(\text{e) }\left(x^3+2x\right)\left(x^2+4\right)=\left(x^2+6x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+2x^2+4x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[\left(x^2+2x^2\right)+\left(4x^2+8\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[x^2\left(x^2+2\right)+4\left(x^2+2\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+4\right)\left(x^2+2\right)\left(3-2x\right)\\ \Leftrightarrow x=3-2x\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\\ \text{Vậy }S=\left\{1\right\}\)
f) Kiểm tra lại hạng tử thứ 2 ở vế phải.
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Mik mới làm có bằng này bạn xem còn căc ý còn lại mik sẽ có làm.
\(a,x^4-16x^2+32x-16=0\)
\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x^2+4x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)
câu nào dễ xơi trước
g) \(x^3+3x^2-2x-6=0\Leftrightarrow x^2\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\x=-3\end{matrix}\right.\)
kl: ...........