4.x+1=21
(-3).x-1=8
(-1005).(x+2)=0
(8+x).(6-x)=0
9.2^x.(2^1005-2^1002+.....+2^3-1)=2^1008-1 *
A=2^1005-2^1002+...+2^3-1 2^3 .
A=2^2.(2^1005-2^1002+......+2^3-1) 8.
A=2^1008-2^1005+.....+2^6-2^3
A=2^1005-2^1002+.....+2^3-1 8.A+A=2^1008-1 9.
A=2^1008-1
Thay 9.a=2^1008-1 vào * ta có: 2^x.(2^1008-1)=2^1008-1
2^x=(2^1008-1):(2^1008-1)
2^x=1
2^x=2^0
x=0
vậy x=0
Lập đề bài cho bài toán này
Tìm x:
a) -152-(3x+1) = (-2).(-77)
b) (-1005).(x+2) = 0
c) (8+x). (6-x) = 0
d) | 4-5x| = 24; x ,x<0
e) (2 +x) mũ 2 = 9
1)4x-20=0 ; 2) 5x+15=0 ; 3) 3x-5=7x+2 ; 4) 4x-(x-1)=2(1+x) ; 5) x2 -2x=0 ; 6) 2(3x-5)-3(x-2)=3(x+4) ; 7) (x+3)(2x-7)=0
8) 5x(x-3)+2x-6=0 ; 9) (3x-1)(2x-1)-(3x-1)(x+2)=0
10)|2x-1|+1=8 ; 11) |x-2|=3x+1 ; 12) |2x|=21-x
Giải các phương trình nha mọi người ^_^
1. |48 - 3x| = 0
2. |-x - 7|= 24
3. |4 - x| = 21
4. |x + 8| + 12 = 0
5. 6 - |x| = 2
Giải:
a) \(\left|48-3x\right|=0\)
\(\Leftrightarrow48-3x=0\)
\(\Leftrightarrow3x=48\)
\(\Leftrightarrow x=\dfrac{48}{3}=16\)
Vậy x = 16.
b) \(\left|-x-7\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24\\-x-7=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=31\\-x=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-31\\x=17\end{matrix}\right.\)
Vậy \(x=-31\) hoặc \(x=17\).
c) \(\left|4-x\right|=21\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21\\4-x=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=25\end{matrix}\right.\)
Vậy \(x=-17\) hoặc \(x=25\).
d) \(\left|x+8\right|+12=0\)
\(\Leftrightarrow\left|x+8\right|=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12\\x+8=-\left(-12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-20\\x=4\end{matrix}\right.\)
Vậy \(x=-20\) hoặc \(x=4\).
e) \(6-\left|x\right|=2\)
\(\Leftrightarrow\left|x\right|=6-2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-4\).
Chúc bạn học tốt!
1. |48 - 3x| = 0.
\(\Leftrightarrow\) 48 - 3x = 0.
\(\Leftrightarrow\) 3x = 48.
\(\Leftrightarrow\) x = \(\dfrac{48}{3}=16.\)
Vậy x = 16.
2. |-x - 7| = 24.
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24.\\-x-7=-24.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=31.\\-x=-17.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
3. |4 - x| = 21.
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21.\\4-x=-21.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
4. |x + 8| + 12 = 0.
|x + 8| = 0 - 12.
|x + 8| = -12.
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=-\left(-12\right).\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=12.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
5. 6 - |x| = 2.
|x| = 6 - 2.
|x| = 4.
\(\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
1. Ta có 2 trường hợp:
TH1: 48-3x=0 <=> 3x=48 <=> x=16
TH2:-48-3x=0
tìm x để:
1) 21 ( x-3) < 0
2) (- 2) (x+7) < 0
3) ( x2 + 1)(x - 5) < 0
4) -3 (8-x) (x2+ 7) không dương
5) (x + 3) ( x - 4) không âm
6) (x - 1) (x +2) < 0
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
1. Tìm x:a) -5(x-1)+13=2(8-x)-15
b) 3.5x+2+4.5x-3=19.510
c) 6.8x-1+8x+1=821+819.6
2. Rút gọn
(-1/2)3-(3/4)3.(-2)2/ 2.(-1)7+(3/4)2-3/8
3) Cho A=1/4x(x-3). tìm x để
a) A=0
b) A > hoặc= 0
c) A<0
tìm x:
a) |4 - x| + 2x = 3
b) |x - 7| + 2x + 5 = 6
c) 3x - |2x + 1| = 2
d) |x + 2| - x = 2
e) |x - 3| = 21
f) |2x + 3| - |x - 3| = 0
g) |x + 1/8| + |x + 2/8| + |x + 5/8| = 4x
h) |x - 2| - |2x + 3| - x = -2
i) |2x - 3| - x = |2 - x|
k) 2. |x - 3| - |4x -1 = 0
a) \(\left|4-x\right|+2x=3\)
<=> \(\left|4-x\right|=3-2x\)
<=> \(\orbr{\begin{cases}4-x=3-2x\left(x\le4\right)\\x-4=3-2x\left(x>4\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\left(tm\right)\\3x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\left(ktm\right)\end{cases}}\)
Vậy x = -1
b) \(\left|x-7\right|+2x+5=6\)
<=> \(\left|x-7\right|=1-2x\)
<=> \(\orbr{\begin{cases}x-7=1-2x\left(đk:x\ge7\right)\\x-7=2x-1\left(đk:x< 7\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=8\\x=-6\left(tm\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{8}{3}\left(ktm\right)\\x=-6\left(tm\right)\end{cases}}\)
Vậy x = -6
c) \(3x-\left|2x+1\right|=2\)
<=> \(\left|2x+1\right|=3x-2\)
<=> \(\orbr{\begin{cases}2x+1=3x-2\left(đk:x\ge-\frac{1}{2}\right)\\2x+1=2-3x\left(đk:x< -\frac{1}{2}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\5x=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\left(ktm\right)\end{cases}}\)
Vậy x = 3
d) \(\left|x+2\right|-x=2\)
<=> \(\left|x+2\right|=x+2\)
<=> \(\orbr{\begin{cases}x+2=x+2\left(đk:x\ge-2\right)\\x+2=-x-2\left(x< -2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\\2x=-4\end{cases}}\)
<=> 0x = 0 (luôn đúng) và x = -2 (ktm)
Vậy x \(\ge\)-2
e) \(\left|x-3\right|=21\)
<=> \(\orbr{\begin{cases}x-3=21\\3-x=21\end{cases}}\)
<=> \(\orbr{\begin{cases}x=24\\x=-18\end{cases}}\)
Vậy x = 24 hoặc x = -18
f) \(\left|2x+3\right|-\left|x-3\right|=0\)
<=> \(\left|2x+3\right|=\left|x-3\right|\)
<=> \(\orbr{\begin{cases}2x+3=x-3\\2x+3=3-x\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\3x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=0\end{cases}}\)
Vậy x thuộc {-6; 0}
g) Ta có: \(\left|x+\frac{1}{8}\right|\ge0\forall x\)
\(\left|x+\frac{2}{8}\right|\ge0\forall x\)
\(\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VT = \(\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\forall x\)
=> VP \(\ge0\) => \(4x\ge0\) => \(x\ge0\)
Do đó: \(x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)
<=> \(3x+1=4x\) <=> \(x=1\left(tm\right)\)
Vậy x = 1
h) \(\left|x-2\right|-\left|2x+3\right|-x=-2\)
<=> \(\left|x-2\right|-\left|2x+3\right|=x-2\)(*)
Lập bảng xét dấu:
x -3/2 2
x - 2 2 - x | 2 - x 0 x - 2
2x + 3 -2x - 3 0 2x + 3 | 2x + 3
Xét x < -3/2 => pt (*) trở thành: 2 - x + 2x + 3 = x - 2
<=> x + 5 = x - 2 <=> 0x = -7 (vô lí)
Xét -3/2 \(\le\) x < 2 => pt (*) trở thành: 2 - x - 2x - 3 = x - 2
<=> 4x = 1 <=> x = 1/4 ((tm)
Xét x \(\ge\) 2 => pt (*) trở thành x - 2 - 2x - 3 = x - 2
<=> 2x = -3 <=> x = -3/2 (ktm)
Vậy x = 1/4
i) |2x - 3| - x = |2 - x|
<=> |2x - 3| - |2 - x| = x (*)
Lập bảng xét dấu
x 3/2 2
2x - 3 3 - 2x 0 2x - 3 | 2x - 3
2 - x 2 - x | 2 - x 0 x - 2
Xét x < 3/2 => pt (*) trở thành: 3 - 2x - 2 + x = x
<=> 2x = 1 <=> x = 1//2 ((tm)
Xét \(\frac{3}{2}\le x< 2\)=> pt (*) trở thành: 2x - 3 - 2 + x = x
<=> 2x = 5 <=> x = 5/2 (ktm)
Xét x \(\ge\)2 ==> pt (*) trở thành: 2x - 3 - x + 2 = x
<=> 0x = -5 (vô lí)
Vậy x = 1/2
k) 2|x - 3| - |4x - 1| = 0
<=> 2|x - 3| = |4x - 1|
<=> \(\orbr{\begin{cases}2\left(x-3\right)=4x-1\\2\left(x-3\right)=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x-6=4x-1\\2x-6=1-4x\end{cases}}\)
<=> \(\orbr{\begin{cases}2x=-5\\6x=7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{7}{6}\end{cases}}\) Vậy ...
1) 4- |x-5|=0
2)|7-3x|=1
3)5-3|x-3 |=x
4) 4-|x-1|=2(x+2)
5)|x+3|-3|x-1|=0
6)11|x+1|-8|x-2|=0
7)2|x-3|=8
8) 2-|x-5|2 =16
Ta có: 4 - |x - 5| = 0
=> | x - 5 | = 4
<=> x - 5 = 4
x - 5 = -4
<=> x = 4 + 5
x = -4 + 5
<=> x = 9
x = 1
1) x=9 hoặc x=1
2)x=2 hoặc x=8/3
3)x=6 hoặc x=2
các bạn giup mình làm hết mình cảm ơn nha ...... ^^