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Bài 1 :
a) \(x^3-9x-28=0\)
\(\Leftrightarrow x^3-4x^2+4x^2-16x+7x-28=0\)
\(\Leftrightarrow x^2\left(x-4\right)+4x\left(x-4\right)+7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x^2+4x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2+4x+4+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(chon\right)\\\left(x+2\right)^2=-3\left(loai\right)\end{matrix}\right.\)
b) Sửa đề nhé
\(\dfrac{x-4}{1004}+\dfrac{x-5}{1003}+\dfrac{x-2}{1006}=3\)
\(\Leftrightarrow\dfrac{x-4}{1004}-1+\dfrac{x-5}{1003}-1+\dfrac{x-2}{1006}-1=0\)
\(\Leftrightarrow\dfrac{x-1008}{1004}+\dfrac{x-1008}{1003}+\dfrac{x-1008}{1006}=0\)
\(\Leftrightarrow\left(x-1008\right)\left(\dfrac{1}{1004}+\dfrac{1}{1003}+\dfrac{1}{1006}\right)=0\)
Vì \(\dfrac{1}{1004}+\dfrac{1}{2003}+\dfrac{1}{2006}>0\)
\(\Leftrightarrow x-1008=0\Leftrightarrow x=1008\)
c) \(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}=\dfrac{3}{4}\)ĐKXĐ : \(x\ne0;-2;-4;-6\)
\(\Leftrightarrow\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{3\cdot2}{4}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+6}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x+6-x}{x\left(x+6\right)}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{6}{x^2+6x}=\dfrac{6}{4}\)
\(\Leftrightarrow x^2+6x=4\)
\(\Leftrightarrow x^2+6x-4=0\)
\(\Leftrightarrow x^2+2\cdot x+3+3^2-13=0\)
\(\Leftrightarrow\left(x+3\right)^2=\left(\pm\sqrt{13}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{13}-3\\x=-\sqrt{13}-3\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
Vậy....
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
Câu 1) Ta có\(a^3+2b^2-4b+3=0\Leftrightarrow a^3=-2.\left(b-1\right)^2-1\)\(\le-1\Rightarrow a^3\le-1\Rightarrow a\le-1\Rightarrow a^2\ge1\)
\(\Rightarrow\hept{\begin{cases}a^2\ge1\\a^2b^2\ge b^2\end{cases}}\)\(\Rightarrow a^2+a^2b^2-2b\ge1+b^2-2b\)\(\Leftrightarrow\left(b-1\right)^2\le0\)
Mà \(\left(b-1\right)^2\ge0\)với mọi b nên \(\left(b-1\right)^2=0\)\(\Rightarrow b=1\)
Thay b=1 vào 2 pt ban đầu được \(\hept{\begin{cases}a^3+2-4+3=0\\a^2+a^2-2=0\end{cases}}\)<=> a=1(tm)
Vậy (a,b)=(1;1)
Câu 2 bạn xem ở đây nhé http://olm.vn/hoi-dap/question/716469.html
Bài làm:
a) đkxđ: \(x\ne\pm1\)
Ta có:
\(M=\frac{x+1}{x^2-1}-\frac{x^2+2}{x^3-1}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{1}{x-1}-\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x+1}{x^2+x+1}\)
\(M=\frac{x^2+x+1-x^2-2-\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x-1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(M=\frac{x\left(1-x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{x}{x^2+x+1}\)
b) Mà x khác 1
=> x = -2, khi đó:
\(M=-\frac{-2}{4-2+1}=\frac{2}{3}\)