tìm x :
\(\frac{5}{8}.8^{x+2}-\frac{3}{5}.8^x=\frac{5}{3}.8^{11}-\frac{3}{5}.8^9\)
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a) Đk: x \(\ne\)-2
Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)
<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10
<=> -4x - 8 = 5x + 10
<=> -4x - 5x = 10 + 8
<=> -9x = 18
<=> x = -2 (ktm)
=> pt vô nghiệm
b) Đk: x \(\ne\)2; x \(\ne\)-3
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
<=> x + 3 - 6x + 12 = -5
<=> -5x = -5 - 15
<=> -5x = -20
<=> x = 4
vậy S = {4}
c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11
Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)
<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)
Vậy S = {0}
\(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{\frac{2}{9}+\frac{2}{5}-\frac{2}{11}}{\frac{8}{5}+\frac{8}{9}-\frac{8}{11}}\)
\(x:8=\frac{2\left(\frac{1}{9}+\frac{1}{5}-\frac{1}{11}\right)}{8\left(\frac{1}{5}+\frac{1}{9}-\frac{1}{11}\right)}\)
\(x:8=\frac{1}{4}\)
\(x=2\)
Vậy..........
1. \(x=\frac{61}{42}\)
2. \(x=\frac{-36}{5}\)
3. \(x=\frac{13}{11}\)
4. \(x=\frac{1}{12}\)
5.\(x=\frac{-5}{2}\)
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)
a) \(\Leftrightarrow2.\left(\frac{2.3^x}{3}+3^x.3^2\right)=2.3^6\left(2+3^3\right)\)
\(\Leftrightarrow2.\left(\frac{2.3^x+3.3^x.3^2}{3}\right)=2.3^6.29\)
\(\Leftrightarrow2.\left[\frac{3^x.\left(2+3.3^2\right)}{3}\right]=2.3^6.19\)
\(\Leftrightarrow2.3^{x-1}.29=2.3^6.29\Leftrightarrow3^{x-1}.29=\frac{2.3^6.29}{2}=3^6.29\Leftrightarrow3^{x-1}=\frac{3^6.29}{29}=3^6\)
\(\Leftrightarrow3^{x-1}=3^6\Leftrightarrow x-1=6\Leftrightarrow x=6+1=7\)
vậy x=7 . Chọn mình nha
mấy bài sao tương tự nếu ko biết thì nhắn tin mình chỉ típ nha