bài 88: tìm x
a) \(5^2.7^3.11^2.x-5^2.7^2.11^4=0\)
b) \(5^2.7^3.11^2.x+5^3.7^2.11=0\)
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Ta có: 52.73.112.x - 52.72.114 = 0
=> 52.72.112(7x - 112) = 0
=> 7x - 121 = 0
=> 7x = 121
=> x = 121 : 7
=> x = 121/7
ta có \(5^2.7^3.11^2.x+5^3.7^2.11=0< =>5^2.7^2.11\left(77x+1\right)=0\)
<=> \(77x+1=0< =>x=-\frac{1}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
=>\(5^2.7^2.11.\left(77x+5\right)=0\)
=>\(77x+5=0\)
=>77x=-5
=>\(x=-\dfrac{5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(5^2.7^2.11.\left(77x+5\right)=0\)
\(77x+5=0\)
\(77x=-5\)
\(x=\dfrac{-5}{77}\)
\(5^2.7^3.11^2.x+5^3.7^2.11=0\)
\(\Leftrightarrow5^2.7^2.11\left(7.11.x+5\right)=0\)
\(\Leftrightarrow77x+5=0\)
\(\Leftrightarrow77x=-5\)
\(\Leftrightarrow x=-\frac{5}{77}\)
1. Tìm x, biết:
a) \(\frac{4}{x}=\frac{8}{6}\). Ta có: \(\frac{8}{6}=\frac{8:2}{6:2}=\frac{4}{3}\Rightarrow x=3\)
b) \(\frac{3}{x-5}=\frac{4}{x+2}\). Ta có: \(5-2=3\)
\(\Rightarrow x=\left(3.5\right)+\left(4.2\right)+3=15+8+3=26\)
c) \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x=\left(-8\right):\left(-2\right)=4\)
2. Rút gọn
a) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}\Leftrightarrow\frac{2^3.2^1.5^2.11.11.7}{2^3.5^2.5^1.7.7.11}\Leftrightarrow\frac{2^1.11}{5^1.7}=\frac{22}{35}\)
b) Tương tự
c) Tương tự
\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4.6}{8}=3\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow3x+4x=20-6\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
\(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x^2=\left(-8\right)\left(-2\right)=16\Rightarrow x=\pm4\)
Ta có:
\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
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A.\(\frac{2^3.9^2}{2^2.3^2.5}=\frac{2^2.2.\left(3^2\right)^2}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
B.\(\frac{2^4.5^5.11^2.7}{2^3.5^3.7^2.11}=\frac{2^3.2.5^3.5^2.11.11.7}{2^3.5^3.7.7.11}=\frac{2.5^2.11}{7}=\frac{61}{7}\)
Trả lời:
a, 52 . 73 . 112 . x - 52 . 72 . 114 = 0
=> 52 . 72 . 112 . ( 7x - 112 ) = 0
=> 7x - 112 = 0
=> 7x = 121
=> x = \(\frac{121}{7}\)
Vậy x = \(\frac{121}{7}\)
b, 52 . 73 . 112 . x + 53 . 72 . 11 = 0
=> 52 . 72 . 11 . ( 7 . 11 . x + 5 ) = 0
=> 77x + 5 = 0
=> 77x = -5
=> x = \(\frac{-5}{77}\)
Vậy x = \(\frac{-5}{77}\)