3x(25+15)-35(5x+3)
giup minh vs
K
Khách
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Giups mk vs thank cacs bn
5 tháng 2 2021
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
R
5 tháng 2 2021
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
5 tháng 6 2019
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
NN
3
TN
18 tháng 7 2017
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
NH
1
TS
21 tháng 1 2016
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
17 tháng 12 2017
\(\frac{x^2}{5x+25}-\frac{10-2x}{x}+\frac{5x+50}{5x+x^2}=\frac{x^2}{5\left(x+5\right)}-\frac{10-2x}{x}+\frac{5x+50}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}-\frac{5\left(x+5\right)\left(10-2x\right)}{5x\left(x+5\right)}+\frac{5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
NN
1
0
26 tháng 3 2018
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
KT
12 tháng 2 2018
a) bạn nhóm 2 cái cuối thành 1 nhóm, 2 cái ở giữa thành 1 nhóm, rồi đặt ẩn phụ là ra
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\) ta có:
\(t\left(t+2\right)-24=0\)
\(\Leftrightarrow\)\(t^2+2t-24=0\)
\(\Leftrightarrow\)\(\left(t-4\right)\left(t+6\right)=0\)
đến đây bn thay trở lại rồi tìm nghiệm nhé
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chi tiet ik