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a) Ta có:\(\left(x+y\right)^2=5^2\)(Vì x + y = 5)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(\Leftrightarrow x^2+2.4+y^2=25\)
\(\Leftrightarrow x^2+8+y^2=25\)
\(\Leftrightarrow x^2+y^2=17\)
b) \(\left(x+y\right)^2=3^2\)(Vì x + y = 3)
\(\Leftrightarrow x^2+2xy+y^2=9\)
\(\Leftrightarrow2xy+5=9\)
\(\Leftrightarrow2xy=4\)
\(\Leftrightarrow xy=2\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3\left(5-2\right)=9\)
a) ta có:(x+y)2=x2+2xy+y2=>x2+y2=(x+y)2-2xy
thay x+y=5;xy=4 vào biểu thức ta có:
52-2×4=25-8=17
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Bài 1:
Theo bài ra ta có:
\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)
\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)
\(=25-10y+y^2+25-10x+x^2-4\)
\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)
\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)
\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)
\(=50-50+5^2-4-4\)
\(=25-8=17\)
Vậy giá trị của \(\left(x-y\right)^2\)là 17
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A) \(\left(x+y\right)^2=\left(x-y\right)^2+4xy=5^2+4.3=37\)
B)
a) \(\left(x+3\right)^2-\left(x-2\right)^2=11\)
\(\Leftrightarrow\)\(x^2+6x+9-\left(x^2-4x+4\right)-11=0\)
\(\Leftrightarrow\)\(x^2+6x+9-x^2+4x-4-11=0\)
\(\Leftrightarrow\)\(10x-6=0\)
\(\Leftrightarrow\)\(10x=6\)
\(\Leftrightarrow\)\(x=\frac{3}{5}\)
Vậy...
b) \(25x^2-9=0\)
\(\Leftrightarrow\)\(\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
Vậy...
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a/VT=x5+x^4.y+x^3.y^2+x^2.y^4+x.y^4-x^4.y-x^3.y^2-x^2.y^3-x.y^4-y^5
=x^5-y^5=VP
=>dpcm
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Ta có: \(x+y=5\)
\(\Rightarrow\left(x+y\right)^2=25\)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(\Leftrightarrow x^2+4+y^2=25\)
\(\Leftrightarrow x^2+y^2=21\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2+y^2\right)-xy\left(x+y\right)\)
\(=105-10\)
\(=95\)
\(A=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.2.5=95\)