a)4*x+79*29=79*33
b)83-3*(7+x)=64
c)(x-7)*(2*x-8)=0
d)580:(x-24)=329-150:2
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a, ĐK: \(x\ne24\)
580 :( x -24 ) =329 -150 : 2
<=> 580 :( x -24 ) = 329 - 75
<=> 580 :( x -24 ) = 254
<=> x - 24 = \(\frac{290}{127}\)
<=> x = \(\frac{3338}{127}\left(TM\right)\)
Vậy \(x=\frac{3338}{127}\)
b, 7 (x-1 ) +35= 25 + 279 :9
<=> 7x - 7 + 35 = 25 + 31
<=> 7x +28 = 56
<=> 7x = 28
<=> x = 4
Vậy x =4
c,4 ( 2x+ 7 ) -3 (3x -2 ) =24
<=> 8x + 28 - 9x + 6 = 24
<=> 34 - x = 24
<=> x = 10
Vậy x = 10
d,( x-1 ) (x-2) =3(x-1)
<=> ( x-1 ) (x-2) - 3(x-1) = 0
<=> (x- 1)(x - 2 - 3) = 0
<=> (x -1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
e, (x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860
x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
Có tất cả: (79 - 3) : 4 + 1 = 20 số hạng \(\Rightarrow\) có 20x
hay x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
3 + 7 + ... + 79 = (79 + 3) x 20 : 2 = 820
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
\(\Rightarrow\) 20x + 820 = 860
\(\Rightarrow\) 20x = 40
\(\Rightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
1/ -7264 + (1543 + 7264)
=-7264 + 1543 + 7264=1543
2/ (144 – 97) – 144
=144-97-144=-97
3/ (-145) – (18 – 145)(Vì có dấu trừ ở trước ngoặc nên p đổi dấu)
=-145-18+145=-18
4/ 111 + (-11 + 27)
=111-11+27=137
24.x-32.x=145-255:51
16.x-9.x=145-5
16.x-9.x=140
x.(16-9)=140
x.7=140
x=140:7
x=20
580:(x-24)=329-150:2=ko bít
7.(x-4)+35=25+279:9
7.(x-4)+35=25+31
7.(x-4)+35=56
7.(x-4)=56-35
7.(x-4)=21
x-4=21:7
x-4=3
x=3+4
x=7
Ủng hộ nha!
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: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
Bài 2:
a: Ta có: \(2x+79=x+45\)
nên 2x-x=45-79
hay x=-24
b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)
\(\Leftrightarrow8-x-x-14=0\)
\(\Leftrightarrow2x=22\)
hay x=11
a) \(2^4.x-3^2.x=145-\frac{255}{51}\Rightarrow x.\left(2^4-3^2\right)=145-5\)
\(\Rightarrow7.x=140\Rightarrow x=20\)
b) \(\frac{580}{x-24}=329-150.2\Rightarrow\frac{580}{x-24}=329-300\Rightarrow\frac{580}{x-24}=29\Rightarrow x-24=20\)
\(\Rightarrow x=44\)
c) \(7.\left(x-14\right)+35=25+\frac{279}{9}\Rightarrow7.\left(x-14\right)=25+31-35\)
\(\Rightarrow7.\left(x-14\right)=21\Rightarrow x-14=3\Rightarrow x=17\)
a) (25x32) : 8
= 25 x (4 x 8) : 8
= (25 x 4) x (8:8)
= 100 x 1
=100
b, 24 x 11 + 6 x 86 x 4 + 72
= 24 x 11 + (6 x 4) x 86 + 24 x 3
= 24 x 11 + 24 x 86 + 24 x 3
= 24 x (11+86+3)
= 24 x 100
= 2 400
a)4*x+79*29=79*33
4*x+2291=2607
4*x=2607-2291=316
x=316:4=79
b)83-3*(7+x)=64
3*(7+x)=83- 64=19
7+x=19:3=19/3
x=19/3-7=-2/3
dễ vây mà