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a, \(4x\left(x-5\right)+2x\left(8-2x\right)=-3\)
\(\Rightarrow4x^2-20x+16x-4x^2=-3\)
\(\Leftrightarrow-4x=-3\Leftrightarrow x=\dfrac{3}{4}\)
Vậy \(x=\dfrac{3}{4}\)
b, \(2x-5\left(x-7\right)=4\left(3-2x\right)-2\)
\(\Rightarrow2x-5x+35=12-8x-2\)
\(\Rightarrow2x-5x+8x=12-2-35\)
\(\Leftrightarrow5x=-25\Leftrightarrow x=-5\)
Vậy \(x=-5\)
Chúc bạn học tốt!!!
có gì ko hiểu bạn hỏi nhé
\(|2x+1|-|x-1|=3x\left(1\right)\)
Ta có:
\(2x+1=0\Leftrightarrow x=\frac{-1}{2}\)
\(x-1=0\Leftrightarrow x=1\)
Lập bảng xét dấu :
+) Với \(x< \frac{-1}{2}\Rightarrow\hept{\begin{cases}2x+1< 0\\x-1< 0\end{cases}\Rightarrow}\hept{\begin{cases}|2x+1|=-2x-1\\|x-1|=1-x\end{cases}\left(2\right)}\)
Thay (2) vào (1) ta được :
\(\left(-2x-1\right)-\left(1-x\right)=3x\)
\(-2x-1-1+x=3x\)
\(-2x+x-3x=1+1\)
\(-4x=2\)
\(x=\frac{-1}{2}\)( loại )
+) Với \(\frac{-1}{2}\le x< 1\Rightarrow\hept{\begin{cases}2x+1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|2x+1|=2x+1\\|x-1|=1-x\end{cases}\left(3\right)}}\)
Thay (3) vào (1) ta được :
\(\left(2x+1\right)-\left(1-x\right)=3x\)
\(2x+1-1+x=3x\)
\(3x=3x\)( luôn đúng chọn )
+) Với \(x\ge1\Rightarrow\hept{\begin{cases}2x+1>0\\x-1>0\end{cases}\Rightarrow\hept{\begin{cases}|2x+1|=2x+1\\|x-1|=x-1\end{cases}\left(4\right)}}\)
Thay (4) vào (1) ta được :
\(\left(2x+1\right)-\left(x-1\right)=3x\)
\(2x+1-x+1=3x\)
\(2x-x-3x=-1-1\)
\(-2x=-2\)
\(x=1\)( chọn )
Vậy \(\frac{-1}{2}\le x\le1\)
\(\left|2x+1\right|-\left|x-1\right|=3x\Rightarrow\left|2x+1-1+x\right|\ge3x\)
\(\Leftrightarrow\left|3x\right|\ge3x\Rightarrow x\in\left\{x\inℤ|x\le0\right\}\)
a) \(\frac{3}{4}+\frac{1}{4}:x=-3\)
\(\frac{1}{4}:x=-3-\frac{3}{4}\)
\(\frac{1}{4}:x=\frac{-15}{4}\)
\(x=\frac{1}{4}:\frac{-15}{4}\)
\(x=\frac{-1}{15}\)
b) \(x-\frac{1}{2}=2,5-x\)
\(x+x=2,5+\frac{1}{2}\)
\(2x=3\)
\(x=\frac{3}{2}\)
c) \(\left(x+\frac{1}{10}\right)+\left(x+\frac{1}{11}\right)=0\)
\(2x+\frac{21}{110}=0\)
\(2x=\frac{-21}{110}\)
\(x=\frac{-21}{110}:2\)
\(x=\frac{-21}{220}\)
Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)
\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
Ta có: \(0< \frac{16}{17}< 1\)
=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)
=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)
=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
a, \(\left(\frac{1}{3}-\frac{1}{2}\right)^x-1=\frac{1}{36}\)
=> \(\left(\frac{-1}{6}\right)^x=\frac{1}{36}+1\)
=> \(\left(\frac{-1}{6}\right)^x=\frac{37}{36}\)
vì ko có số nào mũ với \(\left(\frac{-1}{6}\right)=\frac{37}{36}\) => x ko tồn tại
b, \(\frac{25}{5}^x=\frac{1}{125}=>5^x=\frac{1}{125}=>5^x=5^{\frac{1}{125}}\)
=> x = \(\frac{1}{125}\)
Bạn ơi đề là \(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\) hay \(\left(\frac{1}{3}-\frac{1}{2}\right)^x-1=\frac{1}{36}\) vậy.
\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow\left(-\frac{1}{6}\right)^{x-1}=\frac{1}{36}\)
\(\Rightarrow\left(-\frac{1}{6}\right)^{x-1}=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)