Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. a, 3x + 2 \(⋮2x-1\)
Có 3(2x - 1) \(⋮2x-1\)
Và 2(3x - 2) \(⋮2x-1\)
=> 6x - 4 - 6x + 3 \(⋮2x-1\)
<=> -1 \(⋮2x-1\)
=> 2x - 1 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> 2x = 2; 0
=> x = 1; 0 (thỏa mãn)
@Lớp 6B Đoàn Kết
1. b, x2 - 2x + 3 \(⋮x-1\)
<=> x(x - 2) + 3 \(⋮x-1\)
<=> x(x - 1) - x + 3 \(⋮x-1\)
<=> x(x - 1) - (x - 1) - 2 \(⋮x-1\)
<=> (x - 1)2 - 2 \(⋮x-1\)
<=> -2 \(⋮x-1\)
=> x - 1 \(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x = 2; 0; 3; -1 (thỏa mãn)
@Lớp 6B Đoàn Kết
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
b: =>3|x-5|=8+4=12
=>|x-5|=4
=>x-5=4 hoặc x-5=-4
=>x=9 hoặc x=1
d: =>2x+6=3-3x-2
=>2x+6=1-3x
=>5x=-5
hay x=-1
e: \(\Leftrightarrow x-3\inƯC\left(70;98\right)\)
\(\Leftrightarrow x-3\in\left\{1;2;7;14\right\}\)
mà x>8
nên \(x\in\left\{10;17\right\}\)
a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)
\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)
Vậy x = \(-6\)
b) \(100=6.7^{\left|x+2\right|}-194\)
\(100+194=6.7^{\left|x+2\right|}\)
\(294=6.7^{\left|x+2\right|}\)
\(294:6=49=7^{\left|x+2\right|}\)
\(\Rightarrow7^2=7^{\left|x+2\right|}\)
\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)
+ x + 2 = -2 \(\Rightarrow\) x = - 4
+ x + 2 = 2 \(\Rightarrow\) x = 0
Vậy x = - 4 hoặc 0
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
\(1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\left(đk:x>2\right)\)
đặt
\(A=1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\)
\(\left(x-1\right)^2A=\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\)
\(\left(x-1\right)^2A-A=\left[\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\right]-\left[1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\right]\)
\(\left[\left(x-1\right)^2-1\right]A=\left(x-1\right)^{2022}-1\)
\(A=\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}\)
\(=>\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\\ =>\left(x-1\right)^{2022}-1=17^{2022}-1\\ =>\left(x-1\right)^{2022}=17^{2022}\\ =>x-1=17\\ =>x=18\left(tm\right)\)