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20 . 2^x + 1 = 10.4^2 + 1
20 . 2^x + 1 = 10 . 16 + 1
20 . 2^x + 1 = 161
20 . 2^x = 161 - 1
20 . 2^x = 160
2^x = 8
2^x = 2^3
=> x = 3
x-1/2=5-x/3.
x-x/3=5+1/2.
2/3*x=11/2.x=11/2:2/3.
x=33/4.
Vậy x=33/4.
\(\frac{3}{x-5}=\frac{-4}{x-2}\)
\(\Leftrightarrow3\left(x-2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x-6=-4x+20\)
\(\Leftrightarrow3x+4x=20+6\)
\(\Leftrightarrow7x=26\)
\(\Leftrightarrow x=\frac{26}{7}\)
Vậy \(x=\frac{26}{7}\)
\(\Rightarrow-4\left(x-5\right)=3\left(x-2\right)\)
\(-4x+20=3x-6\)
\(-4x-3x=-6-20\)
\(-7x=-26\)
\(x=\frac{26}{7}\)
\(-2.\left(-x-5\right)+28=20-3.\left(x+4\right)\)
\(\Rightarrow2x+10+28=20-3x-12\)
\(\Rightarrow2x+3x=20-12-10-28\)
\(\Rightarrow5x=-30\)
\(\Rightarrow x=-6\)
Vậy x = -6
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 \(x\in\varnothing\)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
(-25)-(5.x-3)=4
5.x-3 = -25 - 4
5.x-3=-29
5.x=-26
x=-26/5
x=-5,2
a)=>x-1;x-3 \(\in\)Ư(-5)={-1;-5;1;5}
còn lại thử từng TH nhé
b)\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
c)=>x2-4;x2-19 trái dấu
Ta có:x^2-4-(x^2-19)=x^2-4-x^2+19=15 >0
\(\Rightarrow\orbr{\begin{cases}x^2-4>0\\x^2-19< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x^2>4\\x^2< 19\end{cases}}\)
Ta có:4<x^2<19
=>x^2\(\in\){9;16}
=>x\(\in\){3;4}
\(\frac{3}{x-5}=-\frac{4}{x-2}\)
=> 3(x - 2) = -4(x - 5)
=> 3x - 6 = -4x + 20
=> 3x + 4x = 20 + 6
=> 7x = 26
=> x = 26/7
\(\frac{3}{x-5}=\frac{-4}{x-2}\)
=> 3(x-2) = -4 ( x - 5 )
=> 3x - 6 = -4x + 20
=> 7x = 26
=> x = \(\frac{26}{7}\)
Vậy ...