Câu 2: Tìm x, biết
a) x + 72 = 36
b) 16 x2 = 64
c) (5.x – 2) – 64 = -36
d) (2x-10) . (5 – x) = 0
giúp mình với mọi người
Tìm x, biết
a. (5x-7)2-25x2=32
b.(x-3)2+(x+1)2-5=0
giúp mình nhé
\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)
Tìm x e Z biết
a, | -5 | + | x - 1 | = | 7 |
b, 2 . | 2x - 4 | - | - 4 | = | - 50 |
giúp mình với nha mọi người !
Thanks !
a) Ta có: \(\left|-5\right|+\left|x-1\right|=\left|7\right|\)
\(\Leftrightarrow\left|x-1\right|+5=7\)
\(\Leftrightarrow\left|x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-1\right\}\)
b) Ta có: \(2\cdot\left|2x-4\right|-\left|-4\right|=\left|-50\right|\)
\(\Leftrightarrow4\cdot\left|x-2\right|-4=50\)
\(\Leftrightarrow4\cdot\left|x-2\right|=54\)
\(\Leftrightarrow\left|x-2\right|=\dfrac{27}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{27}{2}\\x-2=-\dfrac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{2}\left(loại\right)\\x=-\dfrac{23}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(x\in\varnothing\)
a, | -5 | + | x-1 | = | 7 |
5 + | x - 1 | = 7
| x - 1 | = 2
TH1 x -1 = 2
x = 3
TH2 x -1 = -2
x= -1
f(x)=-2x+6
f(x)=x2 -6x+5
f(x)=(x+3)(4-x)
f(x)=-x2 +4/x2-2x+1
bài 2 giải bpt sau
a (x-2)(x2+2x-3)>/=0
b x2-9/-x+5<0
giúp mình với ạ
\(a)\left(x-2\right)\left(x^2+2x-3\right)\ge0.\)
Đặt \(f\left(x\right)=\left(x-2\right)\left(x^2+2x-3\right).\)
Ta có: \(x-2=0.\Leftrightarrow x=2.\\ x^2+2x-3=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-3.\end{matrix}\right.\)
Bảng xét dấu:
x \(-\infty\) -3 1 2 \(+\infty\)
\(x-2\) - | - | - 0 +
\(x^2+2x-3\) + 0 - 0 + | +
\(f\left(x\right)\) - 0 + 0 - 0 +
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left[-3;1\right]\cup[2;+\infty).\)
\(b)\dfrac{x^2-9}{-x+5}< 0.\)
Đặt \(g\left(x\right)=\dfrac{x^2-9}{-x+5}.\)
Ta có: \(x^2-9=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-3.\end{matrix}\right.\)
\(-x+5=0.\Leftrightarrow x=5.\)
Bảng xét dấu:
x \(-\infty\) -3 3 5 \(+\infty\)
\(x^2-9\) + 0 - 0 + | +
\(-x+5\) + | + | + 0 -
\(g\left(x\right)\) + 0 - 0 + || -
Vậy \(g\left(x\right)< 0.\Leftrightarrow x\in\left(-3;3\right)\cup\left(5;+\infty\right).\)
Tìm X biết
a)22.x-5=32
b)42x-1=64
c)(2.X+2)2=121
d)2x3-1=255
mn ơi giúp mik với mik cần gấp
\(a,\Rightarrow2^{2x-5}=2^5\Rightarrow2x-5=5\Rightarrow x=5\\ b,\Rightarrow4^{2x-1}=4^3\Rightarrow2x-1=3\Rightarrow x=2\\ c,\Rightarrow\left[{}\begin{matrix}2x+2=11\\2x+2=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\\ d,\Rightarrow2^{x^3}=256=2^8\Rightarrow x^3=8\Rightarrow x=2\)
a)
\(2^{2x-5}=2^5\)
2x-5=5
2x=10
x=5
b)
\(4^{2x-1}=4^3\)
2x-1=3
2x=4
x=2
c)
\(\left(2x+2\right)^2=11^2\)
2x+2=11
2x=9
x=9/2
a) \(2^{2x-5}=32\)
⇒\(2^{2x-5}=2^5\)
⇒\(2x-5=5\)
⇒\(2x=10\)
⇒\(x=5\)
tìm x biết:
a) x - 3= (3 - x)^2
b) x^3 + 3/2x^2 + 3/4x + 1/8 = 1/64
c) 8x^3 - 50x = 0
d) (x - 2) (x^2 + 2x + 7) + 2(x^2 - 4) - 5(x - 2) = 0
e) x(x + 3) - x^2 - 3x = 0
f) x^3 + 27 + (x + 3) (x - 9) = 0
Giúp mik vs ạ
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
thực hiện phép tính
a) 5x(3x2-4x+2)
b)(x2-3x)(3x2-x+4)
2) tìm x biết
a) x3-6x2+12x=0
b)x3+9x2+27x+27=0
giúp mình mn ơi
thực hiện phép tính
a) 5x(3x2-4x+2)
b)(x2-3x)(3x2-x+4)
2) tìm x biết
a) x3-6x2+12x=0
b)x3+9x2+27x+27=0
giúp mình mn ơi
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
a) 16. x2 =64
b) (5.x-2)-64= -36
c) (2x-10).(5-x)=0
a)16. x2 = 64
x2 = 64 : 16
x2 = 4
x2 = 22
⇒ x = 2
b) (5.x - 2) - 64 = -36
(5.x - 2) = -36 + 64
5.x - 2 = 28
5.x = 28 + 2
5.x = 30
x = 30 : 5
x = 6
c) (2x - 10).(5 - x) = 0
TH1: 2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
TH2: 5 - x = 0
x = 5 - 0
x = 5
⇒ Vậy x = 5.
\(16.x^2=64\)
\(=>x^2=64:16\)
\(=>x^2=4\)
\(=>x=2\) hoặc \(x=-2\)
_______
\(\left(5x-2\right)-64=-36\)
\(=>5x-2=\left(-36\right)+64\)
\(=>5x-2=28\)
\(=>5x=28+2\)
\(=>5x=30\)
\(=>x=30:5\)
\(=>x=6\)
_______
\(\left(2x-10\right).\left(5-x\right)=0\)
TH1: \(2x-10=0\)
\(=>2x=0+10\)
\(=>2x=10\)
\(=>x=10:2\)
\(=>x=5\)
TH2: \(5-x=0\)
\(x=5-0\)
\(x=5\)
\(=>x=5\)
bài 1: Tìm x biết
a) 2.x+(-5)=-18
b) 3.3=81
c) 64-4.(5-x)=40
a: =>2x=-18+5=-13
=>x=-13/2
b: =>3^x-1=81
=>x-1=4
=>x=5
c: =>4(5-x)=24
=>5-x=6
=>x=-1