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a, \(\frac{x-4}{12}=\frac{x-6}{11}\)
\(\Rightarrow11\left(x-4\right)=12\left(x-6\right)\)
\(\Rightarrow11x-44=12x-72\)
\(\Rightarrow11x-12x=-72+44\)
\(\Rightarrow-x=-28\)
=> x = 28
b, \(\frac{3x+2}{-7}=\frac{x-4}{-2}\)
\(\Rightarrow-2\left(3x+2\right)=-7\left(x-4\right)\)
\(\Rightarrow-6x+\left(-4\right)=-7x-\left(-28\right)\)
\(\Rightarrow-6x+\left(-4\right)=-7x+28\)
\(\Rightarrow-6x+7x=28-\left(-4\right)\)
\(\Rightarrow x=32\)
c, \(\frac{1-4x}{5}=\frac{2x-1}{-3}\)
\(\Rightarrow-3\left(1-4x\right)=5\left(2x-1\right)\)
\(\Rightarrow-3-\left(-12x\right)=10x-5\)
\(\Rightarrow-3+12x=10x-5\)
\(\Rightarrow-3+5=10x-12x\)
\(\Rightarrow2=-2x\)
=> x = 2 : (-2)
=> x = -1
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\(\frac{x}{3}\)- \(\frac{1}{2}\)= \(\frac{3x}{4}\)
\(\frac{x}{3}\)- \(\frac{2}{4}\) =\(\frac{3x}{4}\)
\(\frac{x}{3}\) = \(\frac{3x}{4}\) + \(\frac{2}{4}\)
\(\frac{x}{3}\) =\(\frac{3x+2}{4}\)
\(\Rightarrow\) 4x = 3(3x + 2)
4x=9x + 6
4x - 9x -6 = 0
-5x - 6 = 0
-5x =6
x= \(\frac{-6}{5}\)
TICK NHÉ !!!!
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a) \(3x-\frac{1}{5}=\frac{4+x}{2}\)
=> \(\frac{15x-1}{5}=\frac{4+x}{2}\)
=> \(\left(15x-1\right).2=\left(4+x\right).5\)
=> \(30x-2=20+5x\)
=> \(30x-5x=20+2\)
=> \(25x=22\)
=> \(x=\frac{22}{25}\)
b) \(\frac{4}{3}x-1=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+4-1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+3}{3}\)
=> \(3x=3\left(4x+3\right)\)
=> \(3x=12x+9\)
=> \(3x-12x=9\)
=> \(-9x=9\)
=> \(x=9:\left(-9\right)=-1\)
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a)
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)
Áp dụng tc của dãy tỉ só bằng nhau
\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)
=> x=2.10=20
y=5.10=50
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)
Mà 2;5 cùng dấu
=> x; y cùng dấu
Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)
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a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
X(3-1/2)=3/4
suy ra X=3/4:5/2=3/10
\(3x-\frac{1}{2}x=\frac{3}{4}\)
\(\Leftrightarrow x\times\left(3-\frac{1}{2}\right)=\frac{3}{4}\)
\(\Leftrightarrow x\times\frac{5}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}\div\frac{5}{2}\)
\(\Leftrightarrow x=\frac{3}{10}\)
Vậy \(x=\frac{3}{10}\)