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TA CÓ : X-2/X-1=X+4/X+7
=> X-2-X-4/X-1-X-7(DÙNG TỈ LỆ THỨC)
= -6/-8
DO ĐÓ: X-2/X-1=-6/-8=6/8
=> 8(X-2)=6(X-1) ( TA NHÂN CHÉO)
=>8X-16=6X-6
=>8X=6X-6+16
=>8X=6X=10
=>8X-6X=10
=>2X=10
=>X=10/2=5
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
=> \(\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
=> \(x^2+7x-2x-14=x^2-x+4x-4\)
=> \(x^2+5x-14=x^2+3x-4\)
=> \(x^2-x^2+5x-3x=-4+14\) (chuyển vế)
=> \(2x=10\Rightarrow x=\frac{10}{2}=5\)
Vậy x = 5
a) Ta có: \(\frac{x-4}{x+1}=\frac{x-15}{x+6}\)
\(\Rightarrow\)\(x^2+6x-4x-24=x^2-15x+x-15\)(nhân chéo)
\(\Rightarrow x^2+2x-24=x^2-14x-15\)
\(\Rightarrow16x=9\)
\(\Rightarrow x=\frac{9}{16}\)
a)Ta có:
\(\frac{x-1}{x+2}=\frac{4}{5}\Leftrightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Leftrightarrow5x-5=4x+8\)
\(\Leftrightarrow5x-4x=8+5\)
\(\Leftrightarrow x=13\)
b)Ta có:
\(2^{2x+1}+4^{x+3}=2^{2x+1}+2^{2x+6}=2^{2x+1}\left(1+2^5\right)=2^{2x+1}.33=264\Leftrightarrow2^{2x+1}=8=2^3\)\(\Rightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)
c)Ta có:
\(\frac{x^2}{-8}=\frac{27}{x}\Leftrightarrow x^3=-8.27=-216\Leftrightarrow x=-6\)
d)Ta có:
\(\frac{x+7}{-20}=\frac{-5}{x+7}\Leftrightarrow\left(x+7\right)^2=\left(-20\right)\left(-5\right)=100\Leftrightarrow\left[{}\begin{matrix}x+7=10\\x+7=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-17\end{matrix}\right.\)e)Ta có:
\(\frac{x}{-8}=\frac{2}{-x^3}\Leftrightarrow x.\left(-x^3\right)=-8.2\)
\(\Leftrightarrow-x^4=-16\Leftrightarrow x^4=16\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(a.\frac{x-1}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{x+2-3}{x+2}=\frac{4}{5}\)
\(\Rightarrow1-\frac{3}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=1-\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{1}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{3}{15}\Rightarrow x+2=15\)
\(\Rightarrow x=13\)( thỏa mãn )
a) Ta có: \(\frac{3}{4}-x=\frac{1}{5}\)
hay \(x=\frac{3}{4}-\frac{1}{5}=\frac{11}{20}\)
Vậy: \(x=\frac{11}{20}\)
b) Ta có: \(\left|x+\frac{2}{5}\right|-\frac{3}{7}=\frac{4}{7}\)
\(\Leftrightarrow\left|x+\frac{2}{5}\right|=\frac{4}{7}+\frac{3}{7}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{5}=1\\x+\frac{2}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1-\frac{2}{5}=\frac{3}{5}\\x=-1-\frac{2}{5}=\frac{-7}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{5};\frac{-7}{5}\right\}\)
c) Ta có: \(\left(x+\frac{1^3}{3}\right):2=\frac{-1}{16}\)
\(\Leftrightarrow x+\frac{1}{3}=\frac{-1}{16}\cdot2=-\frac{1}{8}\)
hay \(x=\frac{-1}{8}-\frac{1}{3}=-\frac{11}{24}\)
Vậy: \(x=\frac{-11}{24}\)
d) Ta có: \(\frac{x+2}{3}=\frac{12}{x+2}\)
\(\Leftrightarrow\left(x+2\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)(tm)
Vậy: \(x\in\left\{4;-8\right\}\)
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
Điều kiện: \(x\ne-1;-4;-7;-10\)
Ta có:
\(\frac{3}{\left(x+1\right)\left(x+4\right)}=\frac{1}{x+1}-\frac{1}{x+4}\)
\(\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{x+4}-\frac{1}{x+7}\)
\(\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{1}{x+7}-\frac{1}{x+10}\)
Vậy:
\(\frac{1}{x+1}-\frac{1}{x+10}=\frac{1}{2}\)
Biến đổi tiếp để tìm x, sau đó đối chiếu với điều kiện khác -1; -4; -7; -11 để loại nghiệm
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x=10\)
\(\Rightarrow x=5\)