bài 1 tìm x biết
a, \(\frac{2}{3}\cdot(3-x)+\frac{1}{2}=\frac{3}{4}\cdot(2\cdot x+1)\)
b, \(\frac{5-x}{3}=\frac{2\cdot x+1}{5}\)
làm nhanh trúng lớn
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b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)
\(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)
Vậy x= 4
a.4^7
b.8^5
c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)
b)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Vậy \(x=-329\)
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
a,
x.2/7.3/4=5/21
x.3/14=5/21
x=5/21:3/14
x=10/9
b,
x.1/2=1/3
x=1/3:1/2
x=2/3
c,
x:4/5=25/8:5/4
x:4/5=5/2
x=5/2.4/5=2
A, x= 5/25 : 3/4 : 2/7 = 14/15
B, x=1/3 : 1/2 = 2/3
C, x=(25/8 : 5/4)x4/5 = 5/2 x 4/5 = 2
\(a,\frac{2}{3}.\left(3-x\right)+\frac{1}{2}=\frac{3}{4}.\left(2.x+1\right) \)
\(2-\frac{2}{3}x+\frac{1}{2}=\frac{3}{2}.\frac{3}{4}x+\frac{3}{4} \)
\(\frac{2}{3}x+2-\frac{1}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{2}{3}x+\frac{3}{2}=\frac{9}{8}x+\frac{3}{4}\)
\(\frac{3}{2}-\frac{3}{4}=\frac{9}{8}x-\frac{2}{3}x\)
\(\frac{6}{4}-\frac{3}{4}=\frac{27}{24}x-\frac{16}{24}x\)
\(\frac{11}{24}x=\frac{3}{4}\)
\(x=\frac{3}{4}:\frac{11}{24}\)
\(x=\frac{3}{4}.\frac{24}{11}\)
\(x=\frac{18}{11}\)
\(Vậy x=\frac{18}{11}\)
\(b,\frac{5-x}{3}=\frac{2x+1}{5}\)
\(\frac{\left(5-x\right).5}{15}=\frac{\left(2x+1\right).3}{15}\)
\(\Rightarrow\left(5-x\right).5=\left(2x+1\right).3\)
\(25-5x=6x+3\)
\(25-3=6x+5x\)
\(\Rightarrow11x=22\)
\(\Rightarrow x=22:11\)
\(\Rightarrow x=2\)
\(Vậy x=2\)