rut gon bieu thuc
(2x-1).(x-2)
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\(2x^2\left(x-2\right)-2x\left(x-1\right)\left(x+1\right)=2x^3-4x^2-2x^3+2x=-4x^2+2x=-2x\left(2x-1\right)\)
\(2x^2\left(x-2\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=2x^3-4x^2-2x\left(x^2-1\right)\)
\(=2x^3-4x^2-2x^3+2x=-4x^2+2x\)
2ax^2-a(1+2x^2)-[a-x(x+a)]
=2ax2-2ax2+a+x2-ax+a
=(2ax2-2ax2)-(a+a)+ax+x2
=0-2a+ax+x2
=x2+ax-2a
Ta có 2x(2x + 1)2 - 3x(x + 3)(x - 3) - 4x(x + 1)2
= 2x(4x2 + 4x + 1) - 3x(x2 - 9) - 4x(x2 + 2x + 1)
= 8x3 + 8x2 + 2x - 3x3 + 27x - 4x3 - 8x2 - 4x
= 8x3 - 3x3 - 4x3 + 8x2 - 8x2 + 2x + 27x - 4x
= x3 + 25x
a oi hinh nhu sai r con +16x2 nua co a , anh tinh lai ho e duoc kh
\(\left(2x+1\right)^2-\left(2x+1\right)\left(2x-1\right)\)
\(=4x^2+4x+1-4x^2+1=4x+2=2\left(x+2\right)\)
chúc bn hc tốt ^^
A2=x+\(\sqrt{2x-1}\)+x-\(\sqrt{2x-1}\)- 2\(\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}\)
A2=2x-2\(\sqrt{x^2-2x+1}\)
A2=2x-2(x-1)=1
=>A=1(vì a>0)
Ta có: \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\) \(\left(ĐK:x\ge\frac{1}{2}\right)\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1\)
\(\Leftrightarrow A\sqrt{2}=2\)
\(\Leftrightarrow A=\sqrt{2}\)
\((2x-1)(x-2)=2x^2-4x-x+2=2x^2-5x+2\)