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Câu c phải là \(\left(\frac{x}{2}-y\right)^3\) chứ không phải \(\left(\frac{4}{2}-2\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a: \(8A=8+8^2+...+8^8\)
\(\Leftrightarrow7A=8^8-1\)
hay \(A=\dfrac{8^8-1}{7}\)
b: \(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=3^{16}-1\)
hay \(B=\dfrac{3^{16}-1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
https://olm.vn/hoi-dap/detail/227952918582.html vào link này xem câu a nha Lê Phương Nhung
b)Q = (x - 1)3 - 4x(x + 1)(x - 1) + 3(x - 1)(x2 + x + 1)
Q = (x - 1)3 - 4x(x2 - 1) + 3(x3 - 1)
Thay x = -2 vào Q ta dc :
(-3)3 - 4 . (-2) . 3 + 3 . (-9) = -27 + 24 - 27 = -30
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Answer:
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^2-1}\) \(ĐK:x\ne1\)
\(\Rightarrow1\left(x^2+x+1\right)+2\left(x-1\right)=3x^2\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Rightarrow x^2+3x-3=3x^2\)
\(\Rightarrow2x^2-3x+1=0\)
\(\Rightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\text{(loại)}\end{cases}}\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\) \(ĐK:x\ne-1;x\ne3\)
\(\Rightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{4x}{2\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Rightarrow2x^2-6x=0\)
\(\Rightarrow2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=3\text{(loại)}\end{cases}}}\)
\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\)
\(\Rightarrow\frac{8-x}{x-7}-\frac{1}{x-7}=8\)
\(\Rightarrow\frac{7-x}{x-7}=8\)
\(\Rightarrow-1=8\)
Vậy phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
a, Với x < 0 pt có dạng
\(1-x+2x=-2\Leftrightarrow x=-3\)(tm)
Với 0 =< x < 1 pt có dạng
\(x-1+2x=-2\Leftrightarrow3x=-1\Leftrightarrow x=-\frac{1}{3}\left(ktm\right)\)
Với x >= 1 pt có dạng \(x-1-2x=-2\Leftrightarrow-x=-1\Leftrightarrow x=1\left(tm\right)\)
b, đk : \(x^3+x+8\ge0\)
TH1 : \(x^3-x-8=x^3+x+8\Leftrightarrow2x=-16\Leftrightarrow x=-8\)(ktmđk)
TH2 : \(8+x-x^3=x^3+x+8\Leftrightarrow2x^3=0\Leftrightarrow x=0\left(tm\right)\)