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Bài 1: Phân tích đa thức thành nhân tử

a) Ta có: \(16x^2-y^2+6y-9\)

\(=16x^2-\left(y^2-6y+9\right)\)

\(=\left(4x\right)^2-\left(y-3\right)^2\)

\(=\left[4x-\left(y-3\right)\right]\left[4x+\left(y-3\right)\right]\)

\(=\left(4x-y+3\right)\left(4x+y-3\right)\)

b) Ta có: \(a^2-16a^2b^2+b^2+2ab\)

\(=\left(a^2+2ab+b^2\right)-\left(4ab\right)^2\)

\(=\left(a+b\right)^2-\left(4ab\right)^2\)

\(=\left(a+b-4ab\right)\left(a+b+4ab\right)\)

c) Ta có: \(x^3-6x^2-9x\)

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

d) Ta có: \(mx^2+my^2-nx^2-ny^2\)

\(=m\left(x^2+y^2\right)-n\left(x^2+y^2\right)\)

\(=\left(x^2+y^2\right)\left(m-n\right)\)

e) Ta có: \(a^3+b^3+a^2c+b^2c-abc\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)

\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)

f) Ta có: \(4x^2-y^2-4x+1\)

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

\(=\left(2x-1\right)^2-y^2\)

\(=\left(2x-1-y\right)\left(2x-1+y\right)\)

g) Ta có: \(\left(2x+3\right)^2+5\cdot\left(2x+3\right)\)

\(=\left(2x+3\right)\left(2x+3+5\right)\)

\(=\left(2x+3\right)\left(2x+8\right)\)

\(=2\left(2x+3\right)\left(x+4\right)\)

h) Ta có: \(3x^2-10x-8\)

\(=3x^2-12x+2x-8\)

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

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

18 tháng 7 2020

cảm ơn nhiều ạ :)))

15 tháng 4 2019

a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)\(\frac{12\left(x+5\right)^2}{4}\)

<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)

<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300

<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0

<=>-124x-297≥0

<=>124x+297≤0

<=>124x≤-297

<=>x≤\(\frac{-297}{124}\)

15 tháng 4 2019

b, Tương tự câu a

c, |5−3x|=2+x

TH1: 5-3x=2+x

<=> -3x - x = 2 - 5

<=> -4x = -3

<=> x = 3/4

TH2: 5-3x = -2 - x

<=> -3x + x = -2 - 5

<=> -2x = -7

<=> x = 7/2

a: \(B=\left(\dfrac{x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{10}{5\left(x+2\right)}+\dfrac{1}{x-2}\right):\dfrac{x^2-4+6-x^2}{x-2}\)

\(=\left(\dfrac{1}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x+2}+\dfrac{1}{x-2}\right):\dfrac{2}{x-2}\)

\(=\dfrac{1-2x+4+x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2}=\dfrac{-x+7}{2\left(x+2\right)}\)

b: Ta có: |x|=1/2

=>x=1/2 hoặc x=-1/2

Thay x=1/2 vào B, ta được:

\(B=\dfrac{-\dfrac{1}{2}+7}{2\left(\dfrac{1}{2}+2\right)}=\dfrac{13}{10}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{\dfrac{1}{2}+7}{2\left(-\dfrac{1}{2}+2\right)}=\dfrac{5}{2}\)

3 tháng 9 2021
Sorry, mình 2k10 ko bt lm
NM
3 tháng 9 2021

ta có ;

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2=\left(x-y+x+y\right)^2-4x^2\)

\(=\left(2x\right)^2-4x^2=0\)

12 tháng 12 2016

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

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

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

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

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

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

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

b,Thayx=2005

\(\Rightarrow A=\frac{2005}{4008}\)

17 tháng 8 2020

Bài làm:

a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)

\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)

\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)

\(A=4+2\sqrt{3}+5\sqrt{3}-1\)

\(A=3+7\sqrt{3}\)

b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)

\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)

\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)

\(A=2\)

17 tháng 8 2020

Phần b mình viết nhầm tên thành A, bn sửa thành B nhé

c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

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

\(C=\sqrt{3}-1-2-\sqrt{3}\)

\(C=-3\)

17 tháng 7 2019

2x^2-4x+x-2-[4x^2-8x+1]

=2x^2-3x-2-4x^2+8x-1

=-2x^2+5x-3

17 tháng 7 2019

(2x + 1)(x - 2) - (2x - 1)2

= (2x + 1)(x - 2) - (4x2 - 4x + 1)

= 2x2 - 3x - 2x - 4x2 + 4x - 1

= -2x2 + x - 3