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a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
13(x+3)+(x+3)(x-3)=6(2x+7)
13x+39+x^2-9-12x-42=0
x^2+x-12=0
x=3 và x=-4
**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(\frac{1}{x^2}+\frac{1}{\left(x+2\right)^2}=\frac{10}{9}\)(ĐKXĐ: \(x\ne0;x\ne-2\) )
\(\Leftrightarrow\frac{\left(x+2\right)^2+x^2}{x^2\left(x+2\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{2x^2+4x+4}{x^4+4x^3+4x^2}=\frac{10}{9}\Rightarrow9\left(2x^2+4x+4\right)=10\left(x^4+4x^3+4x^2\right)\)
\(\Leftrightarrow10x^4+40x^3+40x^2=18x^2+36x+36\)
\(\Leftrightarrow10x^4+40x^3+22x^2-36x-36=0\)
\(\Leftrightarrow10x^3\left(x-1\right)+50x^2\left(x-1\right)+72x\left(x-1\right)+36\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x^3+50x^2+72x+36\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[10x^2\left(x+3\right)+20x\left(x+3\right)+12\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(10x^2+20x+12\right)=0\)
Mà \(10x^2+20x+12=10\left(x+1\right)^2+2>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)(thỏa mãn ĐKXĐ)
Tập nghiệm của pt: \(S=\left\{1;-3\right\}\)
Phân thức cuối hình như mẫu sai rồi bạn
Phải là (x+9)(x+10) mới đúng chứ