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Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x+1}{2}=\frac{y+3}{4}\)\(=\frac{z+5}{6}\)\(=\frac{2.\left(x+1\right)+3.\left(y+3\right)+4.\left(z+5\right)}{2.2+3.4+4.6}\)
\(=\frac{2x+2+3y+9+4z+20}{4+12+24}\)\(=\frac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{40}\)
\(=\frac{9+31}{40}=\frac{40}{40}=1\)
Cứ thế là tìm x+1 rồi tìm x
y+3 y
x+5 z
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(=>\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
\(=>\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)
\(=>\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=-\frac{7}{6}\)
\(=>x=-\frac{7}{6}:\frac{3}{2}=-\frac{7}{9}\)
b) \(\frac{26+x}{39-x}=\frac{6}{7}\)
=> 7( 26+ x) = 6(39-x)
=>182 +7x = 234 - 6x
=> 7x+6x = 234-182
=> 13x= 52
=> x=4
a) \(\frac{26+x}{39+x}=\frac{6}{7}\)
=> 7(26+x) = 6(39+x)
=> 182 + 7 x = 234 + 6x
=> 7x - 6x = 234 - 182
=> x = 52
a) |x - 1,7| = 2,3
Xét 2 trường hợp:
TH1: x - 1,7 = -2,3
x = -2,3 +1,7
x = -0,6
TH2: x - 1,7 = 2,3
x = 2,3 + 1,7
x = 4
Vậy: Tự kl :<
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0
=> x + 100 = 0 => x = -100
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)
\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)
Vì 1/47 + 1/48 - 1/49 khác 0
Nên x -50 = 0 => x = 50
x-3/-3 = -27/x-3
=> (x-3)(x-3)=(-3)(-27)
=> (x-3)^2 = 81=9^2
=> x-3=9 hoặc x-3=-9
=> x=12 hoặc x=-6
Ta có :\(\frac{x-3}{-3}=\frac{-27}{x-3}\)
\(\Rightarrow\left(x-3\right)\left(x-3\right)=\left(-3\right).\left(-27\right)\) ( Tính chất tỉ lệ thức )
\(\Rightarrow\left(x-3\right)^2=81\)
\(\Rightarrow\left(x-3\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow x-3=\pm9\)
\(\Rightarrow x=-6;12\)
Ta có: \(-\frac{2}{5}\le x\frac{-7}{5}< \frac{3}{5}\)
\(\Leftrightarrow\frac{-2}{5}\le\frac{x.\left(-7\right)}{5}< \frac{3}{5}\)
\(\Rightarrow-2\le x\left(-7\right)< 3\)
\(\Rightarrow x=\left(0\right)\)
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
giúp mik với
\(\frac{2-x}{x+3}=\frac{6}{5}\)
<=>\(5\left(2-x\right)=6\left(x+3\right)\)
<=>\(10-5x=6x+18\)
<=>\(\left(-5x\right)-6x=18-10\)
<=>\(-11x=8\)
<=>\(x=\frac{-8}{11}\)