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a) x.21=6.7
x.21=42
x=42:21
x = 2
b) y . 20 = -5.28
y.20 = -140
y = (-140) : 20
y = -7
a)=>x*21=7*6
=>x*21=42
=>x=42/21
x=2
b)=>y*20=(-5)*28
=>y*20=-140
=>y=-140/20
y=-7
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ -\dfrac{2}{3}x+\dfrac{1}{6}-\dfrac{2}{3}x+\dfrac{1}{3}=0\)
\(-\dfrac{4}{3}x+\dfrac{1}{2}=0\\ -\dfrac{4}{3}x=-\dfrac{1}{2}\\ x=\dfrac{3}{8}\)
\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^{x+1}=\dfrac{1}{5}2^7+\dfrac{1}{3}2^8\)
\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^x\cdot2=\dfrac{1}{5}2^7+\dfrac{1}{3}2^7\cdot2\)
\(2^x\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)
\(2^x=2^7\\ x=7\)
a) x=\(\dfrac{5.6}{-10}=-3.\)
b) y=\(\dfrac{3.77}{-33}=-7.\)
x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
Lời giải:
Ta có: \(\frac{2}{x}+\frac{5}{y}-\frac{7}{xy}=1\)
\(\Leftrightarrow \frac{2y+5x}{xy}-\frac{7}{xy}=1\)
\(\Leftrightarrow \frac{2y+5x-7}{xy}=1\Rightarrow 2y+5x-7=xy\)
\(\Rightarrow 2y+5x-xy=7\)
\(\Rightarrow y(2-x)+5x=7\)
\(\Rightarrow y(2-x)-5(2-x)=-3\)
\(\Rightarrow (y-5)(2-x)=-3=1(-3)=(-1).3=3(-1)=(-3).1\)
Lập bảng ta thu được:
\((x=5,y=6)\)
\((x=-1,y=4)\)
\((x=3,y=8)\)
\((x,y)=(1,2)\)