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a: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
b: Sửa đề: \(\left(x^3+6x^2+12x+8\right)+3\left(x^2+4x+4\right)+3\left(x+2\right)\)
\(=x^3+6x^2+12x+8+3x^2+12x+12+3x+6\)
\(=x^3+9x^2+27x+26\)
a) Ta có:
\(x+y=3\)
\(\Rightarrow\left(x+y\right)^2=9\)
\(\Leftrightarrow x^2+2xy+y^2=9\)
\(\Leftrightarrow5+2.xy=9\)
\(\Leftrightarrow2xy=4\)
\(\Rightarrow xy=2\)
Ta có:
\(x^3+y^3=\left(x+y\right).\left(x^2-xy+y^2\right)\)
\(\Rightarrow x^3+y^3=3.\left(5-2\right)\)
\(\Rightarrow x^3+y^3=9\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
1. \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
2. \(\left(a-b\right)^3+3ab\left(a-b\right)\)
\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=a^3-b^3\)
a) \(E=4x^2+y^2-4x-2y+3=\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của E là 1 khi \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)vậy GTNN của E là 1 khi \(x=\dfrac{1}{2};y=1\)
b) \(G=x^2+2y^2+2xy-2y=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)-1\)
\(=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của G là \(-1\) khi \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\) vậy GTNN của G là \(-1\) khi \(y=1;x=-1\)
c) \(H=x^2+14x+y^2-2y+7=\left(x^2+14x+49\right)+\left(y^2-2y+1\right)-43\)
\(=\left(x+7\right)^2+\left(y-1\right)^2-43\ge-43\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của H là \(-43\) khi \(\left\{{}\begin{matrix}\left(x+7\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=1\end{matrix}\right.\) vậy GTNN của H là \(-43\) khi \(x=-7;y=1\)
d) câu này hình như đề sai
\(A=\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(A=\left(A^3+3A^2B+3AB^2+B^3\right)-\left(A^3-3A^2B+3AB^2-B^3\right)-\left(6A^2B\right)\)
\(A=A^3+3^2B+3AB^2+B^3-A^3+3A^2B-3AB^2+B^3-6A^2B\)
\(A=0\)
Các bài tiếp theo làm tương tự nhá
2. Khai triển hđt, nhân phân phối, rút gọn.
3.
\(M=\left(x+y\right)^3+2x^2+4xy+2y^2=\left(x+y\right)^3+2\left(x^2+2xy+y^2\right)=\left(x+y\right)^3+2\left(x+y\right)^2\)
Thay x+y=7 vào bt trên, ta có: \(7^3+2.7^2=441\)
Vậy ...
\(N=\left(x-y\right)^3-x^2+2xy-y^2=\left(x-y\right)^3-\left(x^2-2xy+y^2\right)=\left(x-y\right)^3-\left(x-y\right)^2\)Thay x-y=5 vào bt trên, ta có: \(5^3-5^2=100\)