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![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(3x^2-10x-8\)
\(=3x^2-12x+2x-8\)
\(=3x\left(x-4\right)+2\left(x-4\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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≤\(\frac{-297}{124}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(B=\dfrac{\dfrac{1}{2}+7}{2\left(-\dfrac{1}{2}+2\right)}=\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có ;
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x-y\right)\left(x+y\right)-4x^2=\left(x-y+x+y\right)^2-4x^2\)
\(=\left(2x\right)^2-4x^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{x-2}{2x-2}+\frac{3}{2x-2}-\frac{x+3}{2x+2}\right):\left(-1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x-2}{2\left(x-1\right)}+\frac{3}{2\left(x-1\right)}+\frac{-\left(x+3\right)}{2\left(x+1\right)}\right):\left(-\frac{1}{1}+\frac{-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{-1\left(x+1\right)-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{x^2-x^2+x+3x-2x-6+3+3}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x-1-x+3}{x+1}\right)\)
=\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}:\frac{2}{x+1}\)
\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}.\frac{x+1}{2}\)
\(=\frac{x}{2\left(x-1\right)}\)
b,Thayx=2005
\(\Rightarrow A=\frac{2005}{4008}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x + 1)(x - 2) - (2x - 1)2
= (2x + 1)(x - 2) - (4x2 - 4x + 1)
= 2x2 - 3x - 2x - 4x2 + 4x - 1
= -2x2 + x - 3