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(x+3)(x^2-3x+9)-(54+x^3)=0
=>x^3+27-54-x^3=0
=>-27=0(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'
\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
b tự làm nốt nhé~
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-x^3-54+x\)
\(M=x+27-54\)
\(M=x+27-54\)
\(M=7-27\)
\(M=-20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b. Câu hỏi của Phạm Thị Thùy Linh - Toán lớp 8 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+3^3-x^3-54\)
\(=27-54\)
\(=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)+\left(2x+y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2+4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(8x^2+2y^2\right)\)
\(=\left(2x+y\right)\left(4x+y\right).2xy\)
biểu thức <=> x3-3x2+9x+3x2-9x+27-54=0
<=> x3- 27 =0
<=> x=3
\(\left(x+3\right)\cdot\left(x^2-3x+9\right)-54=0\)
\(x^3+27=54\)
\(x^3=54-27=27\)
\(\Rightarrow x=3\)