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Đàm Tùng Vận
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Nguyễn Lê Phước Thịnh
2 tháng 10 2021 lúc 22:34

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

Lấp La Lấp Lánh
2 tháng 10 2021 lúc 22:34

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

ducquang050607
2 tháng 10 2021 lúc 22:37

a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(2x^2-4xy+\dfrac{15}{4}y^2\)

b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(x^2-4x+4+x^2+6x+9-2x^2+2\)

\(2x+15\)

Lê Đoàn Song Tú
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Nguyễn Lê Phước Thịnh
4 tháng 11 2021 lúc 22:34

\(=\left(x-3\right)\left(x^2+1-x^2+1\right)=2\left(x-3\right)\)

Tường Vy
4 tháng 11 2021 lúc 22:38

(x2 + 1)(x - 3) - (x - 3)(x2 - 1)
= [x2 + 1 - (x2 - 1)](x - 3)

= (x2 + 1 - x2 + 1)(x - 3)

= 2(x - 3)

Hoài An
4 tháng 11 2021 lúc 22:45

`=(x-3)[x^2+1-(x^2-1)]`

`=(x-3)(x^2+1-x^2+1)`

`=2(x-3)`

^($_DUY_$)^
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Nguyễn Lê Phước Thịnh
13 tháng 11 2023 lúc 21:26

a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

Duong Tue Tam
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YangSu
16 tháng 6 2023 lúc 10:34

\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3x+6\sqrt{x}-\left(x-1\right)\)

\(=3x+6\sqrt{x}-x+1\)

\(=2x+6\sqrt{x}+1\)

\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)

\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)

\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)

\(=-x+8\sqrt{x}+1\)

\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)

\(=3x-3\sqrt{x}-2+x-1\)

\(=4x-3\sqrt{x}-3\)

\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(=x-9-\left(2x-3\sqrt{x}-2\right)\)

\(=x-9-2x+3\sqrt{x}+2\)

\(=-x+3\sqrt{x}-7\)

\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)

\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)

\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)

\(=x-4-4x-6\sqrt{x}+4\)

\(=-3-6\sqrt{x}\)

Hoàng Thùy Dương
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Nguyễn Hoàng Minh
6 tháng 10 2021 lúc 7:26

\(=x^3-27-4x^2+4x-1=x^3-4x^2+4x-28\)

lucky tomato
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Nguyễn Việt Hoàng
31 tháng 3 2020 lúc 9:06

\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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Trịnh Xuân Ngọc
1 tháng 4 2020 lúc 9:18

ĐKXD: x\(\ne\)-1,-2,-3

Ta có

\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

Chúc bạn học tốt

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Nobi Nobita
3 tháng 4 2020 lúc 15:52

\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne-2\\x\ne-3\end{cases}}\)

\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}+\frac{-2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+2\right)^2}+\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+3\right)\left(x+2\right)^2}\)

\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{x^2+5x+6-2\left(x^2+4x+3\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)\(=\frac{x^2+5x+6-2x^2-8x-6+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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Đàm Tùng Vận
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Lấp La Lấp Lánh
5 tháng 10 2021 lúc 12:01

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

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êfe
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