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a)x3+3x2+3x+1
=x3+3x2*1+3x*12+13
=(x+1)3
b)(x+y)2-9x2
=y2+2xy+x2-9x2
=y2-2xy+4xy-8x2
=y(y-2x)+4x(y-2x)
=(y-2x)(y+4x)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+3x-10\)
\(=x^2+5x-2x-10\)
\(=\left(x^2+5x\right)-\left(2x+10\right)\)
\(=x\left(x+5\right)-2\left(x+5\right)\)
\(=\left(x-2\right)\left(x+5\right)\)
Thích hđt thì chiều :))
x2 + 3x - 10
= ( x2 + 3x + 9/4 ) - 49/4
= ( x + 3/2 )2 - ( 7/2 )2
= ( x + 3/2 - 7/2 )( x + 3/2 + 7/2 )
= ( x - 2 )( x + 5 )
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\(a)\)
\(4x^2-4x+1\)\(=\left(2x-1\right)^2\)
\(b)\)
\(\left(3x+2\right)\left(2-3x\right)\)\(=4-9x^2\)
\(c)\)
\(\left(x-3\right)\left(x^2+3x+9\right)\)\(=x^3-27\)
\(a,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1=\left(2x-1\right)^2\)
\(b,\left(3x+2\right)\left(2-3x\right)=\left(2+3x\right)\left(2-3x\right)=2^2-\left(3x\right)^2\)
\(c,\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+3x.1+3^2\right)=x^3-3^3\)
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1) \(\left(3x+7\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(3x+7-2x+3\right)\left(3x+7+2x-3\right)=0\)
\(\Leftrightarrow\left(x+10\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+10=0\\5x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-10\\x=\frac{-4}{5}\end{cases}}\)
Vạy ...
phần 2 tương tự áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\((4x-1)^2-(5-3x)^2=0\)
\(\Leftrightarrow(4x-1-5-3x)(4x+1+5-3x)=0\)
\(\Leftrightarrow(x-6)(x+6)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy : ...
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a3 + b3=(a+b)(a2-ab+b2)
(a + b)3 =a3+b3+3ab(a+b)
a2 + b2=a2+2ab+b2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(6x-1\right)^2-\left(3x+2\right)\)
\(=36x^2-12x+1-3x-2\)
\(=36x^2-15x-1\)
bn ktra lại đề nhé
hk tốt
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a) \(\left(3x-4\right)^2+2\left(3x-4\right)\left(2x+4\right)+\left(2x+4\right)^2\)\(=\left(3x-4+2x+4\right)^2=\left(5x\right)^2=25x^2\)
b)\(\left(3x+4\right)^2+\left(7+3x\right)^2-\left(6x+8\right)\left(3x+7\right)\)
\(=\left(3x+4\right)^2-2\left(3x+4\right)\left(7+3x\right)+\left(7+3x\right)^2\)
\(=\left[3x+4-\left(7+3x\right)\right]^2=\left(3x+4-7-3x\right)^2=\left(-3\right)^2=9\)
c)\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(\left(2x\right)^2-1^2\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=14x^2\)
xong rồi đấy,bạn k cho mình nhé
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= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
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a) Đặt a + b = x ; a - b = y. Khi đó:
\(\left(a+b\right)^3-\left(a-b\right)^3\)
\(\Leftrightarrow x^3-y^3\)
\(\Leftrightarrow\left[x-y\right]\left[x^2+xy+y^2\right]\)
Thế lại vào ta có:
\(\Leftrightarrow\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(\Leftrightarrow\left[\left(a-a\right)+\left(b+b\right)\right]\left[\left(a^2+b^2+2ab\right)+\left(a^2-b^2\right)+\left(a^2+b^2-2ab\right)\right]\)
\(\Leftrightarrow2b\left[\left(a^2+a^2+a^2\right)+\left(b^2-b^2+b^2\right)+\left(2ab-2ab\right)\right]\)
\(\Leftrightarrow2b\left[3a^2+b^2\right]\)
Mik làm tuỳ theo mình piết thôi nhé
a) ( a + b )3- ( a - b )3= a3 + b3 - a3 - b3 = a3 - a3 + b3 - b3 = 0
b) tương tự như ở trên!!! Hơi khác một tí!!!
c) ( 6x - 1 )2 - ( 3x + 2 ) = ..........
\(=7^2-2.7.3x+\left(3x\right)^2\)
\(=49-42x+9x^2\)
\(\left(7-3x\right)^2\)
\(=7^2-2.7.3x+\left(3x\right)^2\)
\(=49-42x+9x^2\)