Tìm x:
a, ( x - 4 ) mũ 3 = 8
b, x/2 = 6/10
Tìm x:
a, 4.( x + 41 ) = 7
b, 4. ( x-3 ) = 7 mũ 2 - 1 mũ 10
a. 4.(x+41) = 7
x + 41 = 7 : 4 = 1,75
x = 1,75 - 41 = -39,25
b. 4.(x-3) = 72 - 110 = 49 - 1 = 48
x - 3 = 48 : 4 = 12
x = 12 + 3 = 15
Tìm x:
a, 4.( x + 41 ) = 400
b, 4. ( x-3 ) = 7 mũ 2 - 1 mũ 10
a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)
a/ \(4.\left(x+41\right)=400\)
\(x+41=\dfrac{400}{4}=100\)
\(\Rightarrow x=59\)
b/ \(4.\left(x-3\right)=7^2-1^{10}\)
\(4.\left(x-3\right)=49-1\)
\(4.\left(x-3\right)=48\)
\(x-3=\dfrac{48}{3}=16\)
\(\Rightarrow x=19\)
#AEZn8
a ) 4 . ( x + 41 ) = 400
x + 41 = 400 : 4
x + 41 = 100
x = 100 - 41
x = 59
b ) 4 . ( x - 3 ) = 72- 110
4 . ( x - 3 ) = 49 - 1
4 . ( x - 3 ) = 48
x - 3 = 48 : 4
x - 3 = 12
x = 12 + 3
x = 15
Tìm X:
a, 25 + 3 ( x-8 ) = 106
b, 3 mũ 2 ( x+4 ) - 5 mũ 2 = 5.2 mũ 2
Tìm X:
a, 25 + 3 ( x-8 ) = 106
b, 3 mũ 2 ( x+4 ) - 5 mũ 2 = 5.5 mũ 2
a. 25 + 3(x-8) = 106
3(x-8) = 106 - 25 = 81
x -8 = 81 : 3 = 27
x = 27 +8 = 35
b. 32(x +4) - 52 = 5.52
9(x+4) - 25 = 5.25 = 125
9(x+4) = 125 + 25 = 150(Đề sai)
Tìm X:
a, 25 + 3 ( x-8 ) = 106
b, 3 mũ 2 ( x+4 ) - 5 mũ 2 = 5.2 mũ 2
3.(x-8) = 106-25
3.(x-8) = 81
x-8 = 81:3
x-8 = 27
x=35
\(a,25+3\left(x-8\right)=106\\ 3\left(x-8\right)=81\\ x-8=27\\ x=35\)
32 . (x + 4 ) - 52 = 5 . 22
9 . ( x + 4 ) - 25 = 20
9 . ( x + 4 ) = 20 + 25
9 . ( x + 4 ) = 45
( x + 4 ) = 45 : 9
x + 4 = 5
x = 5 - 4
x = 1
Tìm x:
a)(x-1)^2+x(5-x)=8
b)x^3-3x^2+x-3
c)(12x^4-6x):6x+2x(2+x)(2-x)=7
a, <=> x2 -2x +1 + 5x -x2 =8
<=> 3x +1 =8
<=> 3x = 7
<=> x= 7/3
b, thiếu đề
c, <=> 2x3 -1 + 2x(4 -x2) = 7
<=> 2x3 + 8x -23 = 8
<=> 8x =8
<=> x=1
Tìm x:
a, 4.( x + 41 ) = 7
b, x. ( x-51 ) 2. 2 mũ 3 + 20
a) 4(x+41)=7
x+41=7/4
x=7/4-41
x=157/4
b) có gì đó sai sai-.-
Tìm x:
a) (3-2x)2-(3+2x)2=8
b) 9x5-72x2=0
c) 5x4-8x2-4=0
d) (x3+x2-4x-4) : (x-2)=0
Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c: Ta có: \(5x^4-8x^2-4=0\)
\(\Leftrightarrow5x^4-10x^2+2x^2-4=0\)
\(\Leftrightarrow x^2-2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
Tìm x:
a) (x+2)2+(x-1)2+(x-3)(x+3)-3x2=-8
b) 2021x(x-2020)-x+2020=0
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
Tìm x:
a) (x+2)2+(x-1)2+(x-3)(x+3)-3x2=-8
b) 2021x(x-2020)-x+2020=0
a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)
\(\Rightarrow2x=-4\Rightarrow x=-2\)
b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)