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1. So sánh
a) \(25^{50}\) và \(2^{300}\)
\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)
\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25< 64\) nên \(25^{50}< 64^{50}\)
Vậy \(25^{50}< 2^{300}\)
b) \(625^{15}\) và \(12^{45}\)
\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)
\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)
Vậy \(625^{15}< 12^{45}\)
1.So sánh
a)\(25^{50}\) và \(2^{300}\)
Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)
b)\(625^{15}\) và \(12^{45}\)
Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)
Bài làm
a) 2( x + 1 ) - 4x = 6
=> 2x + 2 - 4x = 6
=> ( 2x - 4x ) + 2 = 6
=> -2x + 2 = 6
=> -2x = 4
=> x = -2
Vậy x = -2
b) 3( 2 - x ) + 4( 5 - x ) = 4
=> 6 - 3x + 20 - 4x = 4
=> ( 6 +20 ) + ( -3x - 4x ) = 4
=> 26 - 7x = 4
=> 7x = 22
=> x = 22/7
Vậy x = 22/7
c) Cũng phân tích như hai câu trên rồi rút gọn ra, sử dụng tính chất phân phối đó, do là phân số nên mik k muốn làm.
d) ( x + 1 )( x - 3 ) = 0
=> \(\hept{\begin{cases}x+1=0\Rightarrow x=-1\\x-3=0\Rightarrow x=3\end{cases}}\)
Vậy x = -1; x = 3
# Học tốt #
Tìm x biết :
a) \(2\left(x+1\right)-4x=6\)
\(\Rightarrow2x+2-4x=6\)
\(\Rightarrow2x-4x=6-2\)
\(\Rightarrow-2x=4\)
\(\Rightarrow x=-2\)
b) \(3\left(2-x\right)+4\left(5-x\right)=4\)
\(\Rightarrow6-3x+20-4x=4\)
\(\Rightarrow-3x-4x=4-6-20\)
\(\Rightarrow-7x=22\)
\(\Rightarrow x=-\frac{22}{7}\)
c) \(\frac{7}{3}.\left(x-\frac{4}{3}\right)+\frac{2}{5}.\left(4-\frac{1}{3}x\right)=0\)
\(\Rightarrow\frac{7}{3}x-\frac{28}{9}+\frac{8}{5}-\frac{2}{15}x=0\)
\(\Rightarrow\left(\frac{7}{3}x-\frac{2}{15}x\right)-\left(\frac{28}{9}-\frac{8}{5}\right)=0\)
\(\Rightarrow\frac{33}{15}x-\frac{68}{45}=0\)
\(\Rightarrow\frac{33}{15}.x=\frac{68}{45}\)
\(\Rightarrow x=\frac{68}{45}:\frac{33}{15}\)
\(\Rightarrow x=\frac{68}{99}\)
d) \(\left(x+1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
a, \(\left(x+1\right)^2=169\)
\(\left(x+1\right)^2=13^2\)
\(x+1=13\)
\(x=13-1\)
\(x=12\)
1.
a) \(\left(x+1\right)^2=169\)
⇒ \(x+1=\pm13\)
⇒ \(\left[{}\begin{matrix}x+1=13\\x+1=-13\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=13-1\\x=\left(-13\right)-1\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=12\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{12;-14\right\}.\)
b) \(\left(x+3\right)^3=-\frac{1}{27}\)
⇒ \(\left(x+3\right)^3=\left(-\frac{1}{3}\right)^3\)
⇒ \(x+3=-\frac{1}{3}\)
⇒ \(x=\left(-\frac{1}{3}\right)-3\)
⇒ \(x=-\frac{10}{3}\)
Vậy \(x=-\frac{10}{3}.\)
c) \(\left(2x-4\right)^4=\frac{1}{625}\)
⇒ \(2x-4=\pm\frac{1}{5}\)
⇒ \(\left[{}\begin{matrix}2x-4=\frac{1}{5}\\2x-4=-\frac{1}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=\frac{1}{5}+4=\frac{21}{5}\\2x=\left(-\frac{1}{5}\right)+4=\frac{19}{5}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{21}{5}:2\\x=\frac{19}{5}:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=\frac{21}{10}\\x=\frac{19}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{21}{10};\frac{19}{10}\right\}.\)
Còn câu d) bạn làm tương tự như mấy câu trên.
Chúc bạn học tốt!
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
\(a,|x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+3=\frac{1}{2}\\x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-7}{2}\end{cases}}\)
\(b,|2x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{1}{2}\\2x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=\frac{-7}{4}\end{cases}}\)
\(c,2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=\frac{-3}{2}\)
\(d,|2x+3|-1=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{2}\\2x+3=\frac{-3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{4}\\x=\frac{-9}{4}\end{cases}}\)
\(e,|2x+3|+5=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{-9}{2}\)(vô lí)
\(f,4-|2x+3|=1\Leftrightarrow|2x+3|=3\Leftrightarrow\orbr{\begin{cases}2x+3=3\\2x+3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
a) x=7-\dfrac{2}{5}+1,62=8,22x=7−52+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453+51−21=4103
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53+74
x=\dfrac{41}{35}x=3541
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321−75+131−0.25
x=2 \dfrac{223}{364}x=2364223
x=7-\dfrac{2}{5}+1,62=8,22x=7−52+1,62=8,22
b) x=4 \dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4 \dfrac{3}{10}x=453+51−21=4103
c) 2 x-x=\dfrac{3}{5}+\dfrac{4}{7}2x−x=53+74
x=\dfrac{41}{35}x=3541
d) x=3 \dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0.25x=321−75+131−0.25
x=2 \dfrac{223}{364}x=2364223
\(1,\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{4}{3}\\ =>\dfrac{1}{3}:x=\dfrac{4}{3}-\dfrac{2}{3}\\ =>\dfrac{1}{3}:x=\dfrac{2}{3}\\ =>x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\\ 2,\dfrac{4}{5}+\dfrac{1}{3}:x=\dfrac{2}{3}\\ =>\dfrac{1}{3}:x=\dfrac{2}{3}-\dfrac{4}{5}=-\dfrac{2}{15}\\ =>x=\dfrac{1}{3}:\dfrac{-2}{15}=\dfrac{-5}{2}\\ 3,\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{3}{4}\\ =>\dfrac{5}{2}:x=\dfrac{3}{4}-\dfrac{2}{3}\\ =>\dfrac{5}{2}:x=\dfrac{1}{12}\\ =>x=\dfrac{5}{2}:\dfrac{1}{12}=30\)