Tim x, biet :
(x+2)(x^2-2x+4)+x(5-x)(x+5)=0
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NH
0
HK
6
5 tháng 8 2017
Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
TG
0
NX
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
1
28 tháng 11 2016
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
PH
1
NN
30 tháng 4 2020
a) \(5^{x+2}-5^{x-1}=3100\) \(\Leftrightarrow5^x.5^2-5^x:5=3100\)
\(\Leftrightarrow5^x.25-5^x.\frac{1}{5}=3100\)\(\Leftrightarrow5^x.\left(25-\frac{1}{5}\right)=3100\)
\(\Leftrightarrow5^x.\frac{124}{5}=3100\)\(\Leftrightarrow5^x=125=5^3\)\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(x-4\right)\left(2x+3\right)< 0\)
TH1: \(\hept{\begin{cases}x-4>0\\2x+3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\2x< -3\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< \frac{-3}{2}\end{cases}}\)( vô lý )
TH2: \(\hept{\begin{cases}x-4< 0\\2x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\2x>-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x>\frac{-3}{2}\end{cases}}\Leftrightarrow\frac{-3}{2}< x< 4\)
mà x là số nguyên \(\Rightarrow-1< x< 4\)
Vậy \(-1< x< 4\)
F
0
TM
1
19 tháng 2 2017
a, \(\left|2x+1\right|=5\Rightarrow2x+1\in\left\{5;-5\right\}\)
+) Nếu :\(2x+1=5\Rightarrow2x=4\Rightarrow x=4\div2=2\)
+) Nếu : \(2x+1=-5\Rightarrow2x=-6\Rightarrow x=-6\div2=-3\)
Vậy \(x\in\left\{2;-3\right\}\)
b, \(\left|x-4\right|=\left|2-x\right|\)
\(\Rightarrow\left[\begin{matrix}x-4=2-x\\x-4=-\left(2-x\right)\end{matrix}\right.\)
+) Nếu : x - 4 = 2 - x
\(\Rightarrow x+x=2+4\Rightarrow2x=6\Rightarrow x=3\)
+) Nếu : x - 4 = - ( 2 - x )
\(\Rightarrow x-4=-2+x\Rightarrow x-x=-2+4\Rightarrow0=2\) ( loại )
Vậy x = 3 thỏa mãn đề bài
c, \(\left|x-5\right|=2-x\Rightarrow\left|x-5\right|+x=2\)
+) Nếu : \(x< 5\Rightarrow x-5< 5-5\Rightarrow x-5< 0\Rightarrow\left|x-5\right|=-x+5\)
Thay vào đề , ta có :
\(-x+5+x=2\Rightarrow-x+x+5=2\Rightarrow5=2\) ( loại )
+) Nếu : \(x\ge5\Rightarrow x-5\ge5-5\Rightarrow x-5\ge0\Rightarrow\left|x-5\right|=x-5\)
Thay vào đề , ta có :
\(\left(x-5\right)-x=2\Rightarrow x-5-x=2\)
\(\Rightarrow x-x-5=2\Rightarrow-5=2\) ( loại )
Vậy \(x\in\varnothing\)
17 tháng 2 2020
\(2\left(|x-1|+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(TH1:x\ge1\Rightarrow|x-1|=x-1\)
\(\Rightarrow2\left(x-1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(2x-\frac{9}{5}\right)=2x-\frac{2}{5}\Rightarrow4x-\frac{18}{5}=2x-\frac{2}{5}\)
\(\Rightarrow4x-2x=\frac{18}{5}-\frac{2}{5}\Rightarrow2x=\frac{16}{5}\Rightarrow x=\frac{16}{5}:2=\frac{16}{10}=\frac{8}{5}\)
\(TH2:x< 1\Rightarrow|x-1|=-x+1\)
\(\Rightarrow2\left(-x+1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(1-\frac{4}{5}\right)=2x-\frac{2}{5}\Rightarrow2\cdot\frac{1}{5}=2x-\frac{2}{5}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{2}{5}\Rightarrow2x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\Rightarrow x=\frac{4}{5}:2=\frac{4}{10}=\frac{2}{5}\)