(x-2).(y+3)=13
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1) \(x+y=10\) mà \(x=y\) nên: \(x=y=\dfrac{10}{2}=5\)
2) \(2x+3y=180\) mà \(x=y\)
Ta có: \(2y+3y=180\Rightarrow5y=180\Rightarrow y=180:5=36\)
Vậy \(x=y=36\)
3) \(x+y=180\) mà \(x=y\) nên: \(x=y=\dfrac{180}{2}=90\)
4) \(3x+5y=13\) mà \(y=2x\) ta có:
\(3x+5\cdot2x=13\Rightarrow13x=13\Rightarrow x=1\)
\(y=2x=2\cdot1=2\)
Các câu còn lại bạn làm tương tự
1.\(13.87+13.12+13\)
\(=13\left(87+12+1\right)\)
\(=13.100=1300\)
2.Đề sai à ???
3.\(x\left(x+4\right)-x\left(x-6\right)\)
\(=x^2+4x-x^2+6x\)
\(=10x\)
\(=10.123=1230\)
1, \(13.87+13.12+13=13\left(87+12+1\right)=13.100=1300\)
2, bổ sung \(\left(x-3\right)2x+\left(x-3\right)y=\left(x-3\right)\left(2x+y\right)\)
Thay x = 13 ; y = 4 ta được : \(\left(13-3\right)\left(26+4\right)=10.30=300\)
3, \(x\left(x+4\right)-x\left(x-6\right)=x\left(x+4-x+6\right)=10x\)
Thay x = 123 ta được \(1230\)
Ta có: \(x+y=5\)
\(\Rightarrow x^2+2xy+y^2=25\)
\(\Rightarrow2xy=12\)
\(\Rightarrow xy=6\)
Vậy \(x^3+y^3=\left(x+y\right)\left(x^2+y^2+xy\right)\)
\(=5.\left(13+6\right)=95\)
\(a)(2{y^4} - 13{y^3} + 15{y^2} + 11y - 3):({y^2} - 4y - 3)=2y^2-5y+1\)
b) \((5{x^3} - 3{x^2} + 10):({x^2} + 1)=5x-3+\dfrac{-5x+13}{x^2+1}\)
Từ \(x+y=3\Rightarrow\left(x+y\right)^2=9\)
\(\Rightarrow x^2+2xy+y^2=9\Rightarrow13+2xy=9\)
\(\Rightarrow2xy=-4\Rightarrow-xy=2\)
Lại có: \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3\cdot\left(13+2\right)=3\cdot15=45\)
`1716:y=12xx13`
`1716:y=156`
`y=1716:156`
`y=11`
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`y xx 2/3+y xx 1/6=10/3`
`y xx(2/3+1/6)=10/3`
`y xx(4/6+1/6)=10/3`
`y xx5/6=10/3`
`y=10/3:5/6`
`y=10/3xx6/5=4`
1716 : y = 12 \(\times\) 13
1716 : y = 156
y = 1716 : 156
y = 11
\(y\times\dfrac{2}{3}+y\times\dfrac{1}{6}=\dfrac{10}{3}\)
\(y\times\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{10}{3}\)
\(y\times\dfrac{5}{6}=\dfrac{10}{3}\)
\(y=\dfrac{10}{3}:\dfrac{5}{6}\)
\(y=4\)
a) \(y-6:2-\left(48-24.2:6-3\right)=0.\)
\(\Rightarrow y-3-\left(48-8-3\right)=0\)
\(\Rightarrow y-3-37=0\)
\(\Rightarrow y=0+37+3=40\)
Vậy y = 40
b) \(\left(7.13+8.13\right):\left(9\frac{2}{3}-y\right)=39\)
\(\Rightarrow195:\left(\frac{29}{3}-y\right)=39\)
\(\Rightarrow\frac{29}{3}-y=195:39=5\)
\(\Rightarrow y=\frac{29}{3}-5=\frac{14}{3}\)
Vậy y=\(\frac{14}{3}\)
a) y - 6 : 2 - ( 48 - 24 x 2 : 6 - 3 ) = 0
y - 6 : 2 - ( 48 - 48 : 6 - 3 ) = 0
y - 6 : 2 - ( 48 - 8 - 3 ) = 0
y - 6 : 2 - 37 = 0
y - 3 = 0 + 37
y - 3 = 37
y = 37 + 3
y = 40
b) ( 7 x 13 + 8 x 13 ) : ( \(9\frac{2}{3}\) - y ) = 39
( 15 x 13 ) : ( \(9\frac{2}{3}\) - y ) = 39
195 : ( \(9\frac{2}{3}\) - y ) = 39
\(9\frac{2}{3}\) - y = 195 : 39
\(9\frac{2}{3}-y=5\)
\(y=9\frac{2}{3}-5\)
\(y=\frac{14}{3}\)