Tìm x biết:
a) |x|-3/4=5/3
b)|2x-1/3|+5/6=1
c)17/2-|2-x-3/4|=-7/4
d)|2x-1|=(-4)mũ 2
Ai trả lời nhanh cho mik , mik tk cho
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 6:Tìm x, biết:A,1/4x-1/3=-5/9;B,3,5-Ix-1/2I=0,75;C,x-1/x-5=6/7;D,(x-4)mũ 2=25;E,2 mũ x+2 mũ x-4=272;F,(x+1/2)(2/3-2x)=0
1. 2 mũ 2x-1 =128
2. -2 phần 3 nhân giá trj tuyệt đói của -5 = -7 phần 6
3. 1/2x - 4=5/7
4. x mũ 2 - 2 =0
5. -1/4+1/4:x=-5/8
6. x: 3/8+5/8=x
mik cần gấp lắm ạ
a) 3/4 - 5/4 : ( x - 1 ) = 1/2
b) (1/2 - x) mũ 2 - 2 mũ 2 = 12
c) (1/2) mũ 2x-1 = 1/16
cho mik xin lời giải chi tiết ak
`a, 3/4 - 5/4 :(x-1) =1/2`
`=> 5/4:(x-1)= 3/4 -1/2`
`=> 5/4:(x-1)= 3/4 - 2/4`
`=> 5/4:(x-1)= 1/4`
`=> x-1= 5/4 : 1/4`
`=> x-1=5`
`=>x=5+1`
`=>x=6`
__
`(1/2-x)^2 -2^2 =12`
`=> (1/2-x)^2 = 12+4`
`=> (1/2-x)^2= 16`
`=> (1/2-x)^2 =4^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
__
`(1/2)^(2x-1) =1/16`
`=> (1/2)^(2x-1) = (1/2)^4`
`=> 2x-1=4`
`=> 2x=4+1`
`=>2x=5`
`=>x=5/2`
\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)
\(x-1=\dfrac{5}{4}:\dfrac{1}{4}\)
\(x-1=5\)
\(x=6\)
\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)
\(\left(\dfrac{1}{2}-x\right)^2-4=12\)
\(\left(\dfrac{1}{2}-x\right)^2=16\)
\(\left(\dfrac{1}{2}-x\right)^2=4^2hoặc\left(\dfrac{1}{2}-x\right)^2=\left(-4\right)^2\)
\(\dfrac{1}{2}-x=4hoặc\dfrac{1}{2}-x=-4\)
=>1/2 -x =4 1/2 -x= -4
=> x=1/2-4 x=1/2-(-4)
=>x=-7/2 x=9/2
vậy x∈{-7/2 ; 9/2}
\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)
\(=>\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^4\)
\(=>2x-1=4\)
\(=>2x=5\)
\(=>x=\dfrac{5}{2}\)
Tìm x:
a) |x|-3/4=5/3
b) |2x-1/3|+5/6=1
c) 17/2-|2x-3/4|=-7/4
d) |2x-1|=(-4) mũ 2
d) khó nhất mk làm nhé :
\(\left|2x-1\right|=\left(-4\right)^2\)
\(\left|2x-1\right|=16\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=16\\2x-1=-16\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=17\\2x=-15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{17}{2}\\x=-\frac{15}{2}\end{cases}}\)
Tìm x ( bài tập xoắn 3 đại số 8 )
1. 25x mũ 2 - 20x + 4 = 0
2. ( 2x - 3 ) mũ 2 - ( 2x + 1 ) ( 2x - 1 ) = 0
3. ( 1/2x - 1 ) ( 1/2x + 1 ) - ( 1/2x - 1 ) mũ 2 = 0
4. ( 2x - 3 ) mũ 2 + ( 2x + 5 ) mũ 2 = 8 ( x + 1 ) mũ 2
5. 4x mũ 2 + 12x -7 = 0
6. 1/4x mũ 2 + 2/3x - 5/9 = 0
7. 24 và 8/9 ( hỗn số ) - 1/4x mũ 2 - 1/3x = 0
Các bạn giúp mik nhé, mik sẽ tick cho các bạn !!!!!!!!!!!!
Tìm x:
1. \(25x^2-20x+4=0\)
⇔ \(\left(5x-2\right)^2=0\)
⇔ \(5x-2=0\)
⇔ \(5x=2\)
⇔ \(x=\dfrac{2}{5}\)
⇒ S = \(\left\{\dfrac{2}{5}\right\}\)
2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)
⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)
⇔ \(4x^2-12x+9-4x^2+1=0\)
⇔ \(-12x+10=0\)
⇔ \(-12x=-10\)
⇔ \(x=\dfrac{5}{6}\)
⇒ S \(=\left\{\dfrac{5}{6}\right\}\)
3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)
⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)
⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)
⇔ \(-2+x=0\)
⇔ \(x=2\)
⇒ S \(=\left\{2\right\}\)
4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)
⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)
⇔ \(8x^2+8x+34=8x^2+16x+8\)
⇔ \(8x+34=16x+8\)
⇔ \(8x-16x=8-34\)
⇔ \(-8x=-26\)
⇔ \(x=\dfrac{13}{4}\)
⇒ S \(=\left\{\dfrac{13}{4}\right\}\)
5.\(4x^2+12x-7=0\)
⇔ \(4x^2+14x-2x-7=0\)
⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)
⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)
⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)
6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)
⇔ \(9x^2+24x-20=0\)
⇔ \(9x^2+30x-6x-20=0\)
⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)
⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)
⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)
7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(896-9x^2-12x=0\)
⇔ \(-896+9x^2+12x=0\)
⇔ \(9x^2+12x-896=0\)
⇔ \(9x^2-84x+96x-896=0\)
⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)
⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)
Tìm X
a)2/7-(17/5+7/4)<<7/2-(1/3-4/5)
b)2-7/5<x-1/3
c)4-4/3<x+1/5<86/7-27/8
d)(-8/5+x):12/13=13/6
e)(2x+3)x(x-2)=0
f)2x^2+4x=0
g)(3+x)x(2x-5)<0
bài hơi dài mà mik ngu toán gặp trúng GVCN thích cho bài cảm phiền vì điều này tuy hơi thất lễ nhưng các bạn giúp mik làm nha không thôi là mik xác định
tìm x:
17–14(x+1)=13–4(x+1)–5(x-3)
7(4x+3)-4(x-1)=15(X+0,75)+7
3x(x+1)−2x(x+2)=x mũ 2 – 1
Bạn nào giúp mik VS ;)
a) 17 - 14( x + 1 ) = 13 - 4( x + 1 ) - 5( x - 3 )
<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15
<=> 17 - 14 - 13 + 4 - 15 = -4x - 5x + 14x
<=> -21 = 5x
<=> x = -21/5
b) 7( 4x + 3 ) - 4( x - 1 ) = 15( x + 0, 75 ) + 7
<=> 28x + 21 - 4x + 4 = 15x + 45/4 + 7
<=> 28x - 4x - 15x = 45/4 + 7 - 21 - 4
<=> 9x = -27/4
<=> x = -3/4
c) 3x( x + 1 ) - 2x( x + 2 ) = x2 - 1
<=> 3x2 + 3x - 2x2 - 4x = x2 - 1
<=> 3x2 + 3x - 2x2 - 4x - x2 = -1
<=> -x = -1
<=> x = 1
a, \(17-14\left(x+1\right)=13-4\left(x+1\right)-5\left(x-3\right)\)
\(\Leftrightarrow17-14x-14=13-4x-4-5x+15\)
\(\Leftrightarrow3-14x=24-9x\Leftrightarrow3-14x-24+9x=0\)
\(\Leftrightarrow-21-5x=0\Leftrightarrow5x=-21\Leftrightarrow x=-\frac{21}{5}\)
b, \(7\left(4x+3\right)-4\left(x-1\right)=15\left(x+0,75\right)+7\)
\(\Leftrightarrow28x+21-4x+1=15x+\frac{45}{4}+7\)
\(\Leftrightarrow9x=-\frac{27}{4}\Leftrightarrow x=-\frac{3}{4}\)
c, \(3x\left(x+1\right)-2x\left(x+2\right)=x^2-1\)
\(\Leftrightarrow3x^2+3x-2x^2-4x=x^2-1\)
\(\Leftrightarrow x^2-x=x^2-1\Leftrightarrow x=1\)
Tìm x , biết:
a) |2x +1| - 3 = x +4
b) |3x - 5 | = 1 - 3x
c) |2x + 2|+|x - 1| = 10
d) |x - 3|+|x +4|+|2x + 6|=10
giúp mik vs ạ,nếu có gì sai xót trong đề thì mn thông cảm ạ
`|2x+1|-3=x+4`
`<=>|2x+1|=x+4+3=x+7(x>=-7)`
`**2x+1=x+7`
`<=>x=7-1=6(tm)`
`**2x+1=-x-7`
`<=>3x=-6`
`<=>x=-2(tm)`
`|3x-5|=1-3x(x<=1/3)`
`**3x-5=1-3x`
`<=>6x=6`
`<=>x=1(l)`
`**3x-5=3x-1`
`<=>-5=-1` vô lý
`|2x+2|+|x-1|=10`
Nếu `x>=1`
`pt<=>2x+2+x-1=10`
`<=>3x+1=10`
`<=>3x=9`
`<=>x=3(tm)`
Nếu `x<=-1`
`pt<=>-2x-2+1-x=10`
`<=>-1-3x=10`
`<=>-11=3x`
`<=>x=-11/3(tm)`
Nếu `-1<=x<=1`
`pt<=>2x+2+1-x=10`
`<=>x+3=10`
`<=>x=7(l)`
Vậy `S={3,-11/3}`
d)
+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\) (Nhận)
+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)
+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)
+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)
Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\)