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(7x-11)3=25.52+200
=> (7x-11)3=800+200
=> (7x-11)3=1000
=> (7x-11)3=103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
(2x-15)5=(2x-15)3
=> (2x-15)5 - (2x-15)3 = 0
=> (2x-15)3 . [ (2x-15)2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}2x=15\\2x=16\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
Mà x thuộc N
=> x = 8
(3x-5)10=(3x-5)9
=> (3x-5)10 - (3x-5)9 = 0
=> (3x-5)9 .[ (3x-5) - 1 ] = 0
=> \(\orbr{\begin{cases}\left(3x-5\right)^9=0\\\left(3x-5\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-5=0\\3x-5=1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=5\\3x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
Mà x thuộc N
=> x = 2
a)(7x-11)^3=1000
(7x-11)^3=10^3
7x-11 =10
7x =10+11=21
x =21:7=3
ta có :
\(\left(7x-11\right)^3=100-73=27=3^3\)
nên ta có
\(7x-11=3\Leftrightarrow7x=14\Leftrightarrow x=2\)
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
(7x-11mũ 3)=2.2.2.2.2.5mũ 90
2.2.2.2.2.5mũ 90 =14400
7x=74
\(30\cdot\left(x+2\right)-6\cdot\left(x-5\right)-24\cdot x=100\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow0x+90=100\)
=> PT vô nghiệm
b) \(|7x+3|=66\)
\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)
\(\text{30.(x+2)-6.(x-5)-24.x=100}\)
\(\Leftrightarrow30x+60-6x+30-24x=100\)
\(\Leftrightarrow90=100\)(vô lí)
Vậy phương trình vô nghiệm
\(|7x+3|=66\)
TH1 \(7x+3=66\Leftrightarrow x=\frac{66-3}{7}=9\)
TH2 \(7x+3=-66\Leftrightarrow x=\frac{-66-3}{7}=-\frac{69}{7}\)
a) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)
b) \(x^{17}=x\Rightarrow x^{17}-x=0\Rightarrow x\left(x^{16}-1\right)=0\Rightarrow x=0\) hay \(x=1\)
c) \(\left(2x-2\right)^3=8\Rightarrow\left(2x-2\right)^3=2^3\Rightarrow2x-2=2\Rightarrow2x=4\Rightarrow x=2\)
d) \(\left(x-6\right)^3=\left(x-6\right)^2\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\Rightarrow\Rightarrow\left(x-6\right)^2\left(x-7\right)=0\)
\(\Rightarrow x-6=0\) hay \(x-7=0\Rightarrow x=6\) hay \(x=7\)
e) \(\left(7x-11\right)^3=2^5.5^2+200\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)
f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\Rightarrow2^{x-1}=24-\left[16-3\right]-3\)
\(\Rightarrow2^{x-1}=24-13-3\Rightarrow2^{x-1}=8=2^3\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)
\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)
\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)
\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)
\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)
\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)
\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-1}{2}\)
\(=\frac{3}{14}+\frac{-7}{14}\)
\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)
\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)
\(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)
\(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)
\(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)
\(\frac{14}{9}\cdot x=-\frac{7}{18}\)
\(x=-\frac{7}{18}:\frac{14}{9}\)
\(x=-\frac{1}{4}\)
`(7x - 11)^3 = 2^5 . 5^2 + 100`
`=> (7x - 11)^3 = 32 . 25 + 100`
`=> (7x - 11)^3 = 800 + 100`
`=> (7x - 11)^3 = 900`
`=> 7x - 11 =` \(\sqrt[3]{900}\)
`=> 7x = 11 +` \(\sqrt[3]{900}\)
`=> x =` \(\dfrac{11+\sqrt[3]{900}}{7}\)
Vậy ...
Sửa đề:
\(\left(7x-11\right)^2=2^5.5^2+100\)
\(\left(7x-11\right)^2=32.25+100\)
\(\left(7x-11\right)^2=800+100\)
\(\left(7x-11\right)^2=900\)
\(\left(7x-11\right)^2=30^2\)
\(\Rightarrow7x-11=30\)
\(7x=30+11\)
\(7x=41\)
\(x=41\div7\)
\(x=\dfrac{41}{7}\)
Vậy \(x=\dfrac{41}{7}\)