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TH
19 tháng 12 2018
Bài 1:
a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)
\(=\dfrac{-5+x}{x\left(x-5\right)}\)
\(=\dfrac{x-5}{x\left(x-5\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)
\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)
\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)
\(=\dfrac{x^3-2x^2-9}{x-3}\)
\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)
\(=\dfrac{x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x^2+x+3\right)}{x-3}\)
\(=x^2+x+3\)
c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3}{x+5}\)
d) Đề sai?
Bài 2:
\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(A=2x+2+9x^2-4-9x^2\)
\(A=2x-2\)
\(A=2\left(x-1\right)\)
Thay x = 15 vào A ta được:
\(A=2\left(15-1\right)\)
\(A=2.14=28\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
d) \(\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)
\(=x^3+2^3\)
\(=x^3+8\)
e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)
\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)
\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)
\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)
10 tháng 7 2023
d: (x+2)(x^2-2x+4)
=(x+2)(x^2-x*2+2^2)
=x^3+8
e: (1/4-x/5)(1/16+x/20+x^2/25)
=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]
=1/64-x^3/125
25 tháng 1 2022
4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)
\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm
5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)
6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)
\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)
10 tháng 4 2022
\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
\(ĐK:x\ne0;-10\)
\(\Leftrightarrow\dfrac{12\left(x+10\right)+12x}{12x\left(x+10\right)}=\dfrac{x\left(x+10\right)}{12x\left(x+10\right)}\)
\(\Leftrightarrow12\left(x+10\right)+12x-x\left(x+10\right)=0\)
\(\Leftrightarrow12x+120+12x-x^2-10x=0\)
\(\Leftrightarrow-x^2+14x+120=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)
Y
10 tháng 4 2022
\(o,\dfrac{x}{2x+6}-\dfrac{x}{2x-2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x+3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x+3\right)-2\left(3x+2\right)}{2\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+x-x^2-3x-6x-4=0\)
\(\Leftrightarrow-8x-4=0\)
\(\Leftrightarrow-4\left(2x+1\right)=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
23 tháng 2 2019
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
20 tháng 1 2022
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
20 tháng 1 2022
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
22 tháng 4 2017
a)
Ta có:
cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
\(=\dfrac{x+12+x-2}{x-2}\cdot\dfrac{x\left(x-2\right)}{\left(5-x\right)\left(5+x\right)}=\dfrac{2x\left(x+5\right)}{\left(5-x\right)\left(5+x\right)}=\dfrac{2x}{5-x}\)