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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
a: \(Q=-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1\)
\(A=x^2y-3x+1-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{2}x^2y-\dfrac{7}{12}xy^2-3x\)
b: \(P=\dfrac{3}{4}xy^2+\dfrac{4}{9}x-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{6}xy^2+\dfrac{16}{9}x-\dfrac{1}{2}x^2y-1\)
\(\frac{x-3}{x+3}=\frac{3}{7}\Leftrightarrow\left(x-3\right).7=3.\left(x+3\right)\)
\(\Leftrightarrow7x-21=3x+9\Rightarrow7x-3x=9-\left(-21\right)\)
\(\Leftrightarrow4x=30;x=7,5\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\Leftrightarrow\left(x+4\right)\left(x+4\right)=5.20\)
\(x^2+\left(4+4\right)^2=100\)
\(x^2=100-64=36\)
\(\Rightarrow x\in\left\{-6;6\right\}\)
a ) x + 5/12 = -2/3
=> x = -2/3 - 5/12
=> x = -8/12 - 5/12
=> x = -13/12
b ) 4/5 + 3/4 : x = 1/2
=> 3/4 : x = 1/2 - 4/5
=> 3/4 : x = 5/10 - 8/10
=> 3/4 : x = -3/10
=> x = 3/4 : -3/10
=> x = -5/2
c ) x/2 + x/3 = 1/4
=> 3x/6 + 2x/6 = 1/4
=> ( 3x + 2x )/6 = 1/4
=> 5x/6 = 1/4
=> 20x/24 = 6/24
=> 20x = 6
=> x = 6 : 20
=> x = 0 , 3
Chúc bạn học giỏi !!!
\(\frac{x-2}{5}+\frac{x-3}{4}=\frac{x-4}{3}+\frac{x-5}{2}\)
=> \(\frac{x-2}{5}-1+\frac{x-3}{4}-1=\frac{x-4}{3}-1+\frac{x-5}{2}-1\)
=> \(\frac{x-7}{5}+\frac{x-7}{4}=\frac{x-7}{3}+\frac{x-7}{2}\)
=> \(\frac{x-7}{5}+\frac{x-7}{4}-\frac{x-7}{3}-\frac{x-7}{2}=0\)
=> \(\left(x-7\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
=> x-7 = 0
=> x= 7
Đề vậy sao giải được bạn ơi
\(\left(x-3\right)^3=\left(x-4\right)^4\)
Để \(\left(x-3\right)^3=\left(x-4\right)^4\)thì
Ta có 2 trường hợp
\(\hept{\begin{cases}x-3=0;x-4=0\left(th1\right)\\x-3=1;x-4=1\left(th2\right)\end{cases}}\)
\(\left(th1\right)\Leftrightarrow\hept{\begin{cases}x-3=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\x=4\end{cases}}}\)
\(\left(th2\right)\Leftrightarrow\hept{\begin{cases}x-3=1\\x-4=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}}\)
Vậy \(x\in\left\{3;4;5\right\}\)