Tìm x, biết:
a) |x| = 4; b) |x| = \(\sqrt 7 \); c) |x+5| = 0; d) \(\left| {x - \sqrt 2 } \right|\) = 0
Bài 4. Tìm số nguyên x , biết:
a) |x - 2|= 0 b) |x + 3|= 1 c) -3 |4 - x|= -9 d) |2x + 1|= -2
Bài 5. Tìm số nguyên x, biết:
a) (x + 3)mũ 2 = 36 b) (x + 5)mũ 2 =100 c) (2x - 4)mũ 2 = 0 d) (x - 1)mũ 3 = 27
Bài 3:
Tìm chữ số x, biết:
a) 9,7x2 < 9,712
b) 8,6x1 > 8,689
Bài 4:
Tìm số tự nhiên x,biết:
a) 0,75 < x < 3,25
b) x < 2,02
Bài 5:
Tìm ba số thập phân thích hợp để viết vào chỗ chấm, sao cho:
0,2 < … < 0,23.
B3:
a) x = 0
b) 9
B4 :
a) x = 1,2,3
b) x = 0,1,2
B5:
0,21; 0,215; 0,22
Tìm x, biết:
a. (3/4)^x= 2^8/3^4
Tìm x biết:
a)x^4+x^3-10x^2+1=(x-2)(x^2+2x+4)
\(\Leftrightarrow x^4+x^3-10x^2+1=x^3-8\)
\(\Leftrightarrow x^4-10x^2+9=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)
Tìm số nguyên x biết:
a)x^2 = 16
b)x^2+(-4) = 140
a: x^2=16
=>x=4 hoặc x=-4
b: =>x^2=144
=>x=12 hoặc x=-12
a)x^2 =16
x =4^2
b)x^2+(-4) = 140
x^2 = 140-(-4)
x^2 = 144
x^2 =12^2
Tìm x, biết:
a) 1/4 x + 7/4 x = -6
b) 1/4x + 2x = 9/2
a) `1/4 x + 7/4 x = -6`
`=> x(1/4 + 7/4) = -6`
`=> x. 2 = -6`
`=> x= -6/2`
`=> x= -3`
Vậy `x= -3`
b) `1/4 x +2x = 9/2`
`=> x(1/4 +2) =9/2`
`=> x(1/4 + 8/4) = 9/2`
`=>x. 9/4 =9/2`
`=> x= 9/2 : 9/4`
`=> x= 2`
Vậy `x=2`
tìm x biết:
a.(x+3)^2-(x+3)(x-3)=0
b.5x(x^2+4)=0
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)
\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)
\(x^2+6x+9-x^2+9=0\)
\(6x+18=0\)
\(6x=-18\)
\(x=-3\)
Vậy x=-3
\(b,5x^3+20x=0\)
\(5x\left(x^2+4\right)=0\)
\(Th1:5x=0=>x=0\)
\(Th2:x^2+4=0\)
\(x^2=-4\)(vô lý)
Vậy x=0
tìm x, biết:
a)(x-3)2 - 4 = 0
b) x2 - 2x = 24
a. (x - 3)2 - 4 = 0
<=> (x - 3)2 - 22 = 0
<=> (x - 3 + 2)(x - 3 - 2) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b. x2 - 2x = 24
<=> x2 - 2x - 24 = 0
<=> x2 - 6x + 4x - 24 = 0
<=> x(x - 6) + 4(x - 6) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\)
TH1:\(x-3=2\text{⇒}x=5\)
TH2:\(x-3=-2\text{⇒}x=1\)
\(a\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2-2^2=0\)
\(\Rightarrow\left(x-3-2\right).\left(x-3+2\right)=0\)
\(\Rightarrow\left(x-5\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{1;5\right\}\)
\(b,x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2+4x-6x-24=0\)
\(\Rightarrow x.\left(x+4\right)-6.\left(x+4\right)\)
\(\Rightarrow\left(x-6\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{-4;6\right\}\)
Hoctot
tìm x biết:
a/2*(x^4+3)-(-9)=17
b/5^2*x+1-3*4^2=-47
2(x4+3)-(9)=17
⇒2x4+6+9=17
⇒2x4+15=17
⇒ 2x4=2
⇒ x4=1
⇒ x=\(\pm1\)
b) 5x2.x+1-3.42=-47
⇒5x3+1-48=-47
⇒5x3-47=-47
⇒5x3=0
⇒x3=0
⇒x=0
a) \(2\left(x^4+3\right)-\left(-9\right)=17\)
\(2x^4+6+9=17\)
\(2x^4=2\)
\(x^4=1\)
⇒ \(x=1\)
a) 2*(x⁴+3)-(-9)=17
=>2x⁴+2*3+9=17
=>2x⁴+6=17-9=8
=>2x⁴=8-6=2
=>x⁴=2/2=1
=>x=1
tìm x, biết:
a, x:3.5=3/4 :-5/6
b,x-1.2/2 = 8/x-1.2
`x :3*5 = 3/4 :(-5/6)`
`x :15 =3/4*(-6/5)=-9/10`
`x = -9/10 *15 =-27/2`
`x-1*2/2 = 8/x -1.2`
`x- 1*1 = 8/x -2`
`x-8/x = -2+1`
`x-8/x =-1`
`x^2 -8x =-x`
`x^2 -8x +x=0`
`x^2 -7x =0`
`x(x-7) =0`
`=>[(x=0),(x=7):}`
`a, x \div 15=-9/10`
`x=-9/10*14`
`x=-27/2`
`b, (x-1*2)/2=8/(x-1*2)`
\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)
`(x-1*2)^2=16`
`(x-1*2)^2=(+-4)^2`
\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)