Rút gọn biểu thức A = |2x + 4| + 2(x - 3) với x > 0.
A. 4x - 2
B. 3 – 4x
C. -10
D. 4x -10
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26 tháng 7 2023
a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 10 2023
a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))
\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)
b) Thay x = 2003 ta có:
\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)
c) \(C>0\) khi:
\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\) (đpcm)
3 tháng 11 2017
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 7 2023
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
KK
28 tháng 8 2021
a. y4 - 14y2 + 49
Gọi y2 là t, ta có:
t2 - 14t + 49
<=> t2 - 14t + 72
<=> (t - 7)2
Thay x2 = t
<=> (x2 - 7)2
b. \(\dfrac{1}{4}-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c. x4 - 16
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
d. x2 - 9
<=> x2 - 32
<=> (x - 3)(x + 3)
Chọn đáp án A