Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, x+15=20-4x
=> x + 4x = 20 - 15
=> 5x = 5
=> x = 5 : 5
=> x = 1
b, 17-x=7-6x
=> -x + 6x = 7 - 17
=> 5x = - 10
=> x = -10 : 5
=> x = -2
c, 4.(x-5)-3.(x+7)= -19
=> 4x - 4.5 - 3.x + 3.7 = -19
=> 4x - 20 - 3x + 21 = -19
=> (4x-3x) + (21-20) = -19
=> 1x + 1 = -19
=> x = -19 - 1
=> x = -18
d, -7.(5-x)-2.(x-10)=15
=> -7x - (-7).5 - 2x - 2.10 = 15
=> -7x - (-35) - 2x - 20 = 15
=> -7x + 35 - 2x - 20 = 15
=> -7x+2x + (35-20) = 15
=> -5x + 15 = 15
=> -5x = 15 - 15
=> -5x = 0
=> x = 0 : (-5)
=> x = 0
e, -5.(2-x)+4.(x-3) = 10.x-15
=> -5x - (-5).2 + 4x - 4.3 = 10x - 15
=> -5x - (-10) + 4x - 12 = 10x - 15
=> -5x + 10 + 4x - 12 = 10x - 15
=> (-5x+4x) - (12-10) = 10x - 15
=> -1x - 2 = 10x - 15
=> -1x - 10x = -15 + 2
=> -11x = -13
=> x = \(\frac{-13}{-11}\)
a.23x +1=128
= 23x x 2=128
=128:2=64=2 mũ 6
vậy x=2
b.(7x-11)3=25+52+200
(7x-11)3=257=6,3579 mũ 3
7x=17,3579
x=2,4797
c.(2x+1)+(2x+2)+(2x+3)+...+(2x+101)=5757
vế trái có 101 số hạng
VT =(2x +1+2x+101).101:2=(4x+102).101:2=5757
(4x+102).101 =5757.2=11514
(4x+102)=11514:101=114
4x=114-102=12
x=12:4=3
vậy x=3
-.-? lớp 6
Có \(\left(5x-7\right)\left(2x+3\right)-\left(7x+2\right)\left(x-4\right)\)
\(=10x^2+15x-14x-21-7x^2+28x-2x+8\)
\(=\left(10x^2-7x^2\right)+\left(15x-14x+28x-2x\right)+\left(-21+8\right)\)
\(=3x^2+27x-13\)
Thay x = 1/2 vào biểu thức đã rút gọn
\(\Rightarrow3\left(\frac{1}{2}\right)^2+27.\frac{1}{2}-13=\frac{5}{4}\)
Vậy giá trị biểu thức là 5/4
(5x-7)(2x+3)-(7x+2)(x-4)
=10x2+15x-14x-21-(7x2-7x+2x-8)
=10x2-x-21-7x2+7x-2x+8
=3x2+4x-13
Thay x=1/2 vào BT ta được:
3*(1/2)2+4*1/2-13
=3/4+2-13
=3/4-11
=-41/4
a) \(3\left(4-2x\right)-2\left(x+3\right)=12-7x\)
\(\Leftrightarrow\)\(12-6x-2x-6=12-7x\)
\(\Leftrightarrow\)\(6-8x=12-7x\)
\(\Leftrightarrow\)\(x=-6\)
Vậy...
b) \(\left|16+\right|3\left(x-2\right)||-5=20\)
\(\Leftrightarrow\)\(\left|16+\right|3\left(x-2\right)||=25\)(1)
Ta thấy: \(\left|3\left(x-2\right)\right|\ge0\)\(\Rightarrow\)\(16+\left|3\left(x-2\right)\right|>0\)
nên từ (1) \(\Rightarrow\) \(16+\left|3\left(x-2\right)\right|=25\)
\(\Leftrightarrow\)\(\left|3\left(x-2\right)\right|=9\)
\(\Leftrightarrow\) \(\orbr{\begin{cases}3\left(x-2\right)=9\\3\left(x-2\right)=-9\end{cases}}\)
\(\Leftrightarrow\) \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
Vậy....
c) \(\left|-5-3^2\right|-||3x+5|-7.2^3|=3^9:3^7\)
\(\Leftrightarrow\)\(14-||3x+5|-56|=9\)
\(\Leftrightarrow\)\(||3x+5|-56|=5\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|-56=5\\\left|3x+5\right|-56=-5\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|=61\\\left|3x+5\right|=51\end{cases}}\)
đến đây bn giải tiếp nhé
2x = (-36)
=> x = (-36) : 2
= > x = (-18)
a.8x+28-9x+6=24
<=> -x+34=24
<=> -x=24-34
<=> x=10
b. 3-6x-x-18=7
<=> -7x=7+18-3
<=> -7x=22
<=> x=22/-7
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
TH1 \(x\ge0;\left|2x-3\right|=2x-3\)
\(2x-3-5=7x+1\)
\(\Leftrightarrow2x-7x=3+5+1=9\)
\(\Leftrightarrow-5x=9\Rightarrow x=-\frac{9}{5}\left(ktm\right)\)
TH2:\(x< 0;\left|2x-3\right|=-\left(2x-3\right)\)
\(-\left(2x-3\right)-5=7x+1\)
\(\Leftrightarrow-2x+3-5=7x+1\)
\(\Leftrightarrow-2x-7x=-3+5+1=3\)
\(\Leftrightarrow-9x=3\Rightarrow x=-\frac{3}{9}=-\frac{1}{3}\left(tm\right)\)
Vậy \(x=-\frac{1}{3}\)