Tìm x biết:\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
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Tim x biet:
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
Tim x biet
c) \(x-3\frac{1}{2}x=-2\frac{6}{7}\)
d) \(-2\frac{1}{3}x-1\frac{3}{4}x+3\frac{2}{3}=3\frac{3}{5}\)
1 tim x \(2014.\left|x-12\right|+\left(x-12\right)^2=2013.\left|12-x\right|\)\(x\)|
2 chung minh \(8^7-2^{18}⋮14\)
3 tim x,y,z biet 4x=7y=3z va x+y+z=61
4 tim a,b,c biet \(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)vs \(a-b=15\)
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câu 1: Câu hỏi của Vương Ái Như - Toán lớp 7 - Học toán với OnlineMath
câu 2:
Ta có: \(8^7-2^{18}=2^{21}-2^{18}=2^{17}.\left(2^4-2\right)=2^{17}.14⋮14\)
câu 3:
\(4x=7y=3x\Rightarrow\frac{4x}{84}=\frac{7y}{84}=\frac{3z}{84}\Rightarrow\frac{x}{21}=\frac{y}{12}=\frac{z}{28}=\frac{x+y+z}{21+12+28}=\frac{61}{61}=1\)
\(\Rightarrow x=21,y=12,z=28\)
câu 4:
\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2.6}=\frac{2b}{3.6}=\frac{3c}{4.6}\Rightarrow\frac{a}{12}=\frac{b}{9}=\frac{c}{8}=\frac{a-b}{12-9}=\frac{15}{3}=5\)
\(\Rightarrow a=5.12=60,b=9.5=45,c=8.5=40\)
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Tim x biet
d) \(\frac{x-1}{21}=\frac{3}{x+1}\)
e) \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\frac{x-1}{21}=\frac{3}{x+1}\)
=> \(\left(x-1\right)\left(x+1\right)=21\cdot3\)
=> \(x^2-1=63\)
=> \(x^2=64\)
=> \(\orbr{\begin{cases}x^2=8^2\\x^2=\left(-8\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
=> \(\frac{12}{13}x=2\)
=> \(x=\frac{13}{6}\)
d, \(\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)
\(\Leftrightarrow x^2-1=63\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)
e, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)
bai 1: Tim x biet
\(\hept{\begin{cases}x-y=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\)
bai 2: Tim x, y biet:
x+\(\left(-\frac{31}{12}\right)^2\)=\(\left(\frac{49}{12}\right)^2\)-x=y2
Bai 9: Tim x,y,z biet:
(x-1)2+(x+y)2+(xy-z)2=0
a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)
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Tim x biet
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
Tim x biet
a)\(\frac{8}{23}.\frac{46}{24}-x=\frac{1}{3}\)
b) \(\frac{10}{12}:x=\frac{28}{9}.\frac{3}{56}\)
c) \(\frac{x-12}{4}=\frac{1}{2}\)
\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)
=> \(\frac{2}{3}-x=\frac{1}{3}\)
=> \(x=\frac{1}{3}\)
\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)
=> \(\frac{10}{12}\div x=\frac{1}{6}\)
=> \(x=\frac{60}{12}=5\)
\(\frac{x-12}{4}=\frac{1}{2}\)
=> \(\left(x-12\right)\cdot2=4\cdot1\)
=> \(2x-24=4\)
=> \(2x=28\)
=> \(x=14\)
Tìm x, biết:
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
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tim x biet
a\(\frac{3}{5}:\frac{7}{x}:\frac{3}{4}=\frac{8}{35}\)
b,\(x< \frac{7}{12}:\frac{4}{5}\)
