giải hộ mk vs (x^2+1)(3x^2-5x+2)=0
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NT
2
PT
9 tháng 9 2018
1) \(2x^4+3x^3-x^2+3x+2=0\)
\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)
\(\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0\)
\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)
\(\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\x^2-x+1=0\end{matrix}\right.\)
Ta có:
\(x^2-x+1\)
\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x
\(\Rightarrow x^2-x+1\) vô nghiệm
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
PT
9 tháng 9 2018
3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)
Đặt x + 3 = a, ta được
\(\left(a-1\right)^4+\left(a+1\right)^4=16\)
\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)
\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)
\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)
\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)
\(\Rightarrow2a^4+8a^2+4a^2+2=16\)
\(\Rightarrow2a^4+12a^2+2-16=0\)
\(\Rightarrow2a^4+12a^2-14=0\)
\(\Rightarrow2a^4-2a^2+14a^2-14=0\)
\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)
\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)
Vì \(a^2\ge0\) với mọi a
\(\Rightarrow a^2+7\ge7\) với mọi a
\(\Rightarrow a^2+7\) vô nghiệm
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+3-1=0\\x+3+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
NC
1
BT
9 tháng 5 2018
x.(2x^2+5x-3)=0
x.(2x^2-x+6x-3)=0
x.(2x-1).(x+3)=0
-> x=0 hoặc x=-3 hoặc x=1/2
6 tháng 3 2020
a) 6x2 - 5x + 3 = 2x - 3x(2 - x)
<=> 6x2 - 5x + 3 = 2x - 6x + 3x2
<=> 6x2 - 5x + 3 = -4x + 3x2
<=> 6x2 - 5x + 3 + 4x - 3x2 = 0
<=> 3x2 - x + 3 = 0
=> Pt vô nghiệm
b) 25x2 - 9 = (5x + 3)(2x + 1)
<=> 25x2 - 9 = 10x2 + 5x + 6x + 3
<=> 25x2 - 9 = 10x2 + 11x + 3
<=> 25x2 - 9 - 10x2 - 11x - 3 = 0
<=> 15x2 - 12 - 11x = 0
<=> 15x2 + 9x - 20x - 12 = 0
<=> 3x(5x + 3) - 4(5x + 3) = 0
<=> (5x + 3)(3x - 4) = 0
<=> 5x + 3 = 0 hoặc 3x - 4 = 0
<=> x = -3/5 hoặc x = 4/3
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 \(x\in\left\{1;\dfrac{5}{2}\right\}\)
NT
5
BA
29 tháng 7 2019
\(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x^2-5x=x-1\)
\(\Leftrightarrow5x^2-5x-x+1=0\)
\(\Leftrightarrow5x^2-6x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-\frac{1}{5}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
BA
29 tháng 7 2019
\(2\left(x-7\right)-x^2+7x=0\)
\(2\left(x-7\right)-x\left(x-7\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
S
8 tháng 4 2020
A. 2x (3x-2) - (3x-2)=0
➜\(\left(2x-1\right)\left(3x-2\right)=0\)
➜\(\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...................................
B. (x+1) (3-x) +x=0
➜\(3x-x^2+3-x+x=0\)
➜\(3x-x^2=0\)
➜\(x\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy.........................V
C. (x-2)^2 = (2x+3)^2
➞\(\left(x-2\right)^2-\left(2x+3\right)^2=0\)
➜\(\left(x-2-2x-3\right)\left(x-2+2x+3\right)=0\)
➜\(\left[{}\begin{matrix}x-2-2x-3=0\\x-2+2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy..................................
D. x^2 -5x+6=0
➜\(x^2-2x-3x+6=0\)
➜\(x\left(x-2\right)-3\left(x-2\right)=0\)
➜\(\left(x-3\right)\left(x-2\right)=0\)
➜\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy.....................................
26 tháng 3 2020
b, - ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Ta có : \(\frac{5x}{x^2-4}-\frac{4}{x+2}=\frac{5}{x-2}\)
=> \(\frac{5x}{x^2-4}-\frac{4\left(x-2\right)}{x^2-4}=\frac{5\left(x+2\right)}{x^2-4}\)
=> \(5x-4\left(x-2\right)=5\left(x+2\right)\)
=> \(5x-4x+8=5x+10\)
=> \(5x-4x-5x=10-8\)
=> \(-4x=2\)
=> \(x=-\frac{1}{2}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-\frac{1}{2}\right\}\)
c, Ta có : \(x^4-15x^2+56=0\)
=> \(\left(x^2\right)^2-\frac{2.x^2.15}{2}+\frac{225}{4}-\frac{1}{4}=0\)
=> \(\left(x^2-\frac{15}{2}\right)^2=\frac{1}{4}\)
=> \(\left[{}\begin{matrix}x^2-\frac{15}{2}=\sqrt{\frac{1}{4}}\\x^2-\frac{15}{2}=-\sqrt{\frac{1}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=\sqrt{\frac{1}{4}}+\frac{15}{2}=8\\x^2=-\sqrt{\frac{1}{4}}+\frac{15}{2}=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\\x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\sqrt{8};-\sqrt{8};\sqrt{7};-\sqrt{7}\right\}\)
26 tháng 3 2020
a)
\(\frac{x-5x-1}{6}=\frac{8-3x}{4}\)
\(\Leftrightarrow\frac{4x-20x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4-48+18x}{24}=0\)
\(\Leftrightarrow\frac{2x-52}{24}=0\)
\(\Rightarrow2x-52=0\)
\(x=\frac{52}{2}=26\)
Anh ko ghi lại đề nha em !
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\3x^2-5x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\x_1=1;x_2=\frac{2}{3}\end{cases}}\)( vn là vô nghiệm nha )
Vậy : x = 1 hoặc x = 2/3
\(\left(x^2+1\right).\left(3x^2-5x+2\right)=0\)
\(x^2\ge0\Rightarrow x^2+1\ge1\)
\(\RightarrowĐể\left(x^2+1\right).\left(3x^2-5x+2\right)=0\)
\(\Rightarrow3x^2-5x+2=0\Rightarrow3x^2-3x-2x+2=0\)
\(\Rightarrow3x.\left(x-1\right)-2.\left(x-1\right)=0\Rightarrow\left(3x-2\right).\left(x-1\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{2}{3}\end{cases}}\)