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Ta có:\(\left(a-b+c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2\\ =2\left(\left(a-b+c\right)^2-\left(b-c\right)^2\right)\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\\ =2\left(a-2b+2c\right)a \)
\(=2a^2-4ab+4ac\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
Thay số từ đề bài vào rùi tính thui :
\(15^2=a^2+2\cdot7+b^2\)
\(\Leftrightarrow225=a^2+b^2+14\)
\(\Leftrightarrow a^2+b^2=225-14=211\)
TK NKA !!!
\(a,\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)
\(=a^2+4x+4-a^2+4\)
\(=4x+8\)
\(=4\left(x+2\right)\)
\(b,\left(a+b\right)^2-\left(a-b\right)^2\)
\(=a^2+2ab+b^2-\left(a^2-2ab+b^2\right)\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)
\(c,\left(3x+4\right)^2-10x-\left(x+4\right)\left(x-4\right)\)
\(=9x^2+24x+16-10x-x^2+16\)
\(=8x^2+14x+32\)
\(=2\left(4x^2+7x+16\right)\)
Bạn ơi bạn có thể ghi câu trả lời ra cụ thể giúp mình có được không ạ ?
`@` `\text {Ans}`
`\downarrow`
\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)
`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`
`= x^2 - xy + xy - y^2 + y^2 - x^2`
`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`
`= 0`
\(2;A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(\frac{1-x}{x+2}\right)\)
\(ĐKXĐ:\hept{\begin{cases}x^2-4\ne0\\1-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne\pm2\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{1-x}\)
\(A=\left(\frac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{1-x}\)
\(A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1-x}=\frac{-6}{\left(x-2\right)\left(1-x\right)}\)
b, Khi x = -4
\(A=\frac{-6}{\left(-4-2\right)\left(1+4\right)}=\frac{-6}{-6.5}=\frac{1}{5}\)
(a^2+b^2+c^2)^2-(a^2+b^2-c^2)
\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2-b^2+c^2\)
\(=2c^2+2ab+2bc+2ac\)
\(=2\left(c^2+ab+bc+ac\right)=2\left[\left(c^2+ac\right)+\left(ab+bc\right)\right]\)
\(=2\left[c\left(a+c\right)+b\left(a+c\right)\right]=2\left(a+c\right)\left(b+c\right)\)
k nha!