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\(\left(x+2\right)^3-x.\left(x+2\right).\left(x-2\right)+6x^2\)
\(=x^3+3x^2.2+3x.2^2+2^3-x.\left(x^2-2^2\right)+6x^2\)
\(=x^3+6x^2+12x+8-\left(x^2-4\right)+6x^2\)
\(=x^3+6x^2+12x+8-x^3+4x+6x^2\)
\(=\left(x^3-x^3\right)+\left(6x^2+6x^2\right)+\left(12x+4x\right)+8\)
\(=12x^2+16x+8\)
mình mới học lớp 7 thui à
Nếu lớp 8 thì sẽ giúp bạn liền
\(\left(3x+4\right)^2-10x-\left(x-4\right)\left(x+4\right)\)
\(=9x^2+24x+16-10x-x^2+16\)
\(=8x^2+14x\)
P = ( 3x + 4 )2 - 10x - ( x - 4 )( x + 4 )
P = 9x2 + 24x + 16 - 10x - ( x2 - 16 )
P = 9x2 + 24x + 16 - 10x - x2 + 16
P = 8x2 + 14x + 32
P = 2( 4x2 + 7x + 16 )
\(\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)
\(=x^2-2x-4x+8-\left(x^2-3x-x+3\right)\)
\(=x^2-2x-4x+8-x^2+3x+x-3\)
\(=-2x+5\)
(x - 4).(x - 2) - (x - 1).(x - 3)
= x2 - 2x - 4x + 8 - ( x2 - 3x - x + 3 )
= x2 - 2x - 4x + 8 - x2 + 3x + x - 3
= 5 - 2x
......
a) P=\(\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\left(x\ne\pm1;x\ne0\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right)\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
P=\(\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x+1}=\frac{x\left(x+1\right)x\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=\frac{x^2}{x-1}\)
vậy P=\(\frac{x^2}{x-1}\left(x\ne\pm1;x\ne0\right)\)
b) ta có \(P=\frac{x^2}{x-1}\left(x\ne\pm1;x\ne0\right)\)
để P<1 => \(\frac{x^2}{x-1}< 1\)
\(\Leftrightarrow\frac{x^2}{x-1}-1< 0\Leftrightarrow\frac{x^2-x+1}{x-1}< 0\Leftrightarrow\frac{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}{x-1}< 0\)
thấy \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
vậy để P-1<0 thì x-1<0
=> x<1. kết hợp với điều kiện ta được \(\hept{\begin{cases}x< 1\\x\ne0\\x\ne-1\end{cases}}\)thì P<1
\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\\ =\left(x+2\right)\left(x+2-x+2\right)\\ =4\left(x+2\right)=4x+8\)
Ta có:(x+2)2-(x+2)(x-2)=(x+2)(x+2-x+2)=4(x+2)