a)(x^4)=x^12/X^5(X khác 0)
b)x^10=25x^8
c)(2x+3)^2=9/121
d)(3x-1)^3=-8/27
a) (x^4)^2=x^12/x^5 (x khác 0 )
b) x^10+=25x^8
c) (2x+3)^2=9/121
d) (3x-1)^3=-8/27
Các bạn giúp mình với
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm
Bài 1: tìm x
1, 2x(3x-1)+1-3x=0
2, x\(^2\)(2x-3)+12-8x=0
3, 25(x-1)\(^2\)-4=0
4, 25x\(^2\)-10x+1=0
5, -4x\(^2\)+\(\dfrac{1}{9}\)=0
6, (x-1)\(^3\)=8
7, (2x-1)\(^3\)+27=0
8, 125+\(\dfrac{1}{8}\)(x-1)\(^3\)=0
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
`@` `\text {Ans}`
`\downarrow`
`5,`
`-4x^2 + 1/9 = 0`
`<=> -4x^2 = 0 - 1/9`
`<=> -4x^2 = -1/9`
`<=> 4x^2 = 1/9`
`<=> x^2 = 1/9 \div 4`
`<=> x^2 = 1/36`
`<=> x^2 = (+-1/6)^2`
`<=> x = +-1/36`
Vậy, `S = {1/36; -1/36}`
`6,`
`(x-1)^3 = 8`
`<=> (x-1)^3 = 2^3`
`<=> x-1=2`
`<=> x = 2 + 1`
`<=> x = 3`
Vậy, `S = {3}`
`7,`
`(2x-1)^3 + 27 = 0`
`<=> (2x - 1)^3 = -27`
`<=> (2x-1)^3 = (-3)^3`
`<=> 2x - 1 = -3`
`<=> 2x = -3 + 1`
`<=> 2x = -2`
`<=> x = -1`
Vậy,` S = {-1}`
`8,`
`125 + 1/8(x-1)^3 = 0`
`<=> 1/8(x-1)^3 = - 125`
`<=> (x-1)^3 = -125 \div 1/8`
`<=> (x-1)^3 = -1000`
`<=> (x-1)^3 = (-10)^3`
`<=> x - 1 = - 10`
`<=> x = -10+1`
`<=> x = -9`
Vậy, `S = {-9}.`
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm
A, (x-2)^2=4x^2+4x+1 B, 25x^2-9=0 C, (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0 D, (2x+1)(x-4)-2(x-3)^2=8 E, (3x-1)^2=(5-x)^2
Tìm x
1. (3x+5)(4-3x)=0
2. 9(3x-2)=x(2-3x)
3. 25x^2 -2=0
4. x^2- 25=6x-9
5. (2x-1)^2-(2x+5)(2x-5)=18
6. x^3-8=(x-2)^3
7. x^3-4x^2+4x=0
8. x^2- 25+2(x+5)=0
9. 2(x^2+8x+16)- x^2+4=0
10. x^2(x-2)+7x=14
(3x+5)(4-3x)=0
3x+5 =0 hoặc 4-3x=0
3x=-5 hoặc 3x=-4
x=-5/3 hoặc x=-4/3
9(3x-2)=x(2-3x)
9(3x-2)-x(3x-2)=0
(3x-2)(9-x)=0
3x-2=0 hoặc 9-x=0
3x=2 hoặc x= -9
x =2/3 hoặc x=-9
vậy x =2/3 ; x= -9
25x^2 - 2=0
(5x)^2 -√2^2=0
(5x-√2)(5x+√2)=0
5x=√2 hoặc 5x = -√2
x=√2/5 hoặc x= -√2/5
vậy x=√2/5 ; x=-√2/5
Tìm x biết
a) (x+2).(x+3) - (x-2).(x+5)=10
b) (3x+2). (2x+9) - (x+2). (8x+11)=(x+1).(3-2x)
c) 3.(2x-1).(3x-1)-(2x-3).(9x-1)=0
d) (5x-8).(4x-5)-(3x-4).(2x+12)=12
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a) -5 (2x + 6 ) +b | 3x + 9 | = 7x
b) 7(x - 4 ) - | 4x - 12 | = 0
c) | x - 5 | - | x - 3 | =0
d)| x + 1| + | -x + 12 | = 2x -7
e) | x + 2| - | x + 6 | - (-15) = 0
f) | 2x- 10 | + | 3y - 4 | = 0
g) | - 3x + 1 | - | 2x + 8 | + 3x = 0
h) 3( x - 1 ) - 5 | x + 3 | = 0
Mọi người giúp mình nhanh nhé ! Mình đang cần gấp
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha