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![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề:
Cách 1:
\(\left(5x-1\right)^2+2.\left(1-5x\right).\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(1-5x\right)^2+2.\left(1-5x\right).\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left(1-5x+5x+4\right)^2\)
\(=5^2\)
\(=25\)
Cách 2:
\(\left(5x-1\right)^2+2.\left(1-5x\right).\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(5x-1\right)^2-2.\left(5x-1\right).\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left(5x-1-5x-4\right)^2\)
\(=\left(-5\right)^2\)
\(=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(5x-4\right)\left(4x-5\right)+\left(5x-1\right)\left(x+4\right)+3\left(3x-2\right)\)
\(=20x^2-41x+20+\left(5x-1\right)\left(x+4\right)+3\left(3x-2\right)\)
\(=20x^2-41+20+5x^2+19x-4+3\left(3x-2\right)\)
\(=20x^2-41x+20+5x^2+19x-4+9x-4\)
\(=25x^2-13x+10\)
2) \(\left(5x-4\right)^2+\left(16-25x^2\right)+\left(5x+4\right)\left(3x+2\right)\)
\(=\left(5x-4\right)^2+16-25x^2+\left(5x-4\right)\left(3x+2\right)\)
\(=25x^2-40x+16^2-25x^2+\left(5x-4\right)\left(3x+2\right)\)
\(=25x^2-40x+16^2-25x^2+15x^2-2x-8\)
\(=15x^2-42x+24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(25x^2-10x+1-25x^2+16=7\)
\(17-10x=7\)
\(10x=10\)
\(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..
Câu sau cũng tương tự nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(P=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=5x-1+2\left(4+5x-20x-25x^2\right)+\left(25x^2+40x+16\right)\)
\(=5x-1+8-30x-50x^2+25x^2+40x+16\)
\(=\left(-50x^2+25x^2\right)+\left(5x-30x+40x\right)+\left(-1+8+16\right)\)
\(=-25x^2+15x+23\)
b, \(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+y^3+3y^2x+3yx^2+x^3+y^3-3y^2x+3yx^2-x^3-3x^2y-3xy^2\)
\(=\left(x^3+x^3-x^3\right)+\left(-y^3+y^3+y^3\right)+\left(-3x^2y+3x^2y+3x^2y-3x^2y\right)+\left(3xy^2+3xy^2-3xy^2-3xy^2\right)\)
\(=x^3+y^3\)
Chúc bạn học tốt!!!
áp dụng hằng đẳng thức ra luôn là (1-5x+5x+4)2=52=25
Ban làm ra luôn đi