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1) \(\frac{x+4}{7+y}=\frac{4}{7}\)\(\Rightarrow7\left(x+4\right)=4\left(7+y\right)\)
\(\Rightarrow7x+28=28+4y\)
\(\Rightarrow7x=4y\)
\(\Rightarrow\frac{x}{4}=\frac{y}{7}\)
áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4}=\frac{y}{7}=\frac{x+y}{4+7}=\frac{22}{11}=2\)
x/4 = 2 => x = 4 x 2 = 8
y/7 = 2 => y = 2 x 7 = 14
a)x-3/x+5=5/7 suy ra 7.(x-3) = 5(x+5)
Tương đương : 7x - 21 = 5x + 25
7x - 5x = 25 + 21 = 46
2x = 46 suy ra : x = 46/2 = 23
Vậy x = 23
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
a) Sửa đề \(\frac{-3}{x+1}=\frac{x+1}{-12}\)
<=> (x + 1)(x + 1) = (-12).(-3)
<=> (x + 1)2 = 36
<=> \(\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
b) \(\frac{x}{5}=-\frac{x+24}{3}\)
=> 3x = -(x + 24).5
<=> 3x = -5x - 120
<=> 8x = -120
<=> x = -15
Vậy x = -15
c) \(\frac{x+2}{x+1}=\frac{x-4}{x-2}\)
<=> \(\frac{x+2}{x+1}-1=\frac{x-4}{x-2}-1\)
<=> \(\frac{1}{x+1}=\frac{-2}{x-2}\)
<=> (x - 2).1 = -2(x + 1)
<=> x - 2 = -2x - 2
<=> 3x = 0
<=> x = 0
Vậy x = 0
d) \(\frac{x+4}{y+7}=\frac{4}{7}\)
<=> \(\frac{x+4}{4}=\frac{y+7}{7}=\frac{x+4+y+7}{4+7}=\frac{x+y+11}{11}=\frac{22+11}{11}=3\)(dãy tỉ số bằng nhau)
<=> \(\hept{\begin{cases}\frac{x+4}{4}=3\\\frac{y+7}{7}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x+4=12\\y+7=21\end{cases}}\Leftrightarrow\hept{\begin{cases}x=8\\y=14\end{cases}}\)
a ) \(-\frac{3}{x+1}=\frac{x+1}{-12}\)
\(\Leftrightarrow\)\(\left(x+1\right).\left(x+1\right)=-3.\left(-12\right)\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=36\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=\pm6\)
\(\Rightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
b ) \(\frac{x}{5}=\frac{x+24}{3}\)
\(\Leftrightarrow\)\(3x=\left(x+24\right).5\)
\(\Leftrightarrow\)\(3x=5x+120\)
\(\Leftrightarrow\)\(-2x=120\)
\(\Leftrightarrow\)\(x=-60\)
d ) \(\frac{x+4}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x+4}{4}=\frac{7+y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+4}{4}=\frac{7+y}{7}=\frac{\left(x+y\right)+\left(4+7\right)}{4+7}=\frac{22+11}{11}=\frac{33}{11}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+4}{4}=3\\\frac{7+y}{7}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+4=12\\7+y=21\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=8\\y=14\end{cases}}\)