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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 \(x\in\varnothing\)
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{9998}{9999}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{9998}{9999}\)
\(\Leftrightarrow5\left(1-\frac{1}{x+1}\right)=\frac{9998}{9999}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{9999}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{9998}{49995}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9998}{49995}=\frac{39997}{49995}\)
\(\Leftrightarrow x=\frac{9998}{39997}\)
a) \(\left(x+5\right).\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0-5\\x=0+4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=+4\end{cases}}}\)
Vậy \(x\in\){-5 ; 4}
b) \(\left(x-5\right)^6=\left(x-5\right)^8\)
Ta có : \(0^n=0\)\(;\)\(1^n=1\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}\Rightarrow\orbr{\begin{cases}x=0+5\\x=1+5\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
Vậy \(x\in\){5 ; 6}
a) (x + 5) . (x - 4) = 0
x + 5 = 0 x = -5
<=> <=>
x - 4 = 0 x = 4
(x-5)^6-(x-5)^4=0
(x-5)^4.[(x-5)^2-1]=0
=>2 trường hợp
(x-5)^4=0=>x-5=0=>x=5
(x-5)^2-1=0 =>(x-5)^2=1=>x-5=-1;1
=>x={4,9}
Vậy x={4,5,9}
\(\left(x+1\right)\left(xy-5\right)=5\)
\(\Rightarrow\left(x+1\right);\left(xy-5\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Xét bảng
x+1 | 1 | -1 | 5 | -5 |
xy-5 | 5 | -5 | 1 | -1 |
x | loại | -2 | 4 | -6 |
y | loại | 0 | 6/4 | -4/6 |
Vậy........................
P/S: ko chắc
\(-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(=>-12x+60+21-7x=5\)
\(=>-19x=5-81=-76\)
\(=>x=\frac{-76}{-19}=4\)
\(\frac{x}{4}=\frac{5}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.5\)
\(\Rightarrow x\left(x+1\right)=4\left(4+1\right)\)
\(\Rightarrow x=4\)
\(\left|x-5\right|=5-x\)
\(\Rightarrow\left[\begin{array}{nghiempt}5-x=x-5\\5-x=-\left(x-5\right)\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}5+5=x+x\\5-x=-x+5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}10=2x\\x\le5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=5\\x\le5\end{array}\right.\)
Vậy \(x=5\) hoặc \(x\le5\) thõa mãn đề bài .
\(\left|x-5\right|=5-x\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-5=5-x\\5-x=5-x\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x\in R\end{array}\right.\)
Vậy x là số thực