Tìm x:
2x+9-2x+8-2x+7-...............-2x+1-2x=1024
Tính:
A=\(\dfrac{125.4^2.\left(2^{15}\right)^3}{13.2^{13}.8^6-64^5}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
25 tháng 12 2023
a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)
31 tháng 12 2023
Bài 1:
a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)
=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)
=>\(2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
=>x=2/2=1
b: \(x^2-2=0\)
=>\(x^2=2\)
=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Bài 2:
a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)
\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)
\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)
\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)\)
\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)
b: A<x<B
=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)
mà \(x\in Z\)
nên \(x\in\left\{-9;-8;...;0;1\right\}\)
DT
Dang Tung
CTVHS
31 tháng 1 2023
\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)
31 tháng 1 2023
\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)
mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)
Khử mẫu ta được :
\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)
do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)
\(Vậy...\)
NT
28 tháng 12 2017
4.
\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)
29 tháng 12 2017
\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)
\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)
\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)
S
14 tháng 1 2019
a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)
\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)
\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)
\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)
24 tháng 8 2021
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
24 tháng 8 2021
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
16 tháng 12 2022
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
20 tháng 1 2022
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
20 tháng 1 2022
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
\(2^{x+9}-2^{x+8}-2^{x+7}-..........-2^{x+1}-2^x=1024\)
\(\Leftrightarrow2^x.2^9-2^x.2^8-.........-2^x.2-2^x=1024\)
\(\Leftrightarrow2^x\left(2^9-2^8-.........-2-1\right)=1024\)
Đặt :
\(A=2^9-2^8-......-2-1\Leftrightarrow2^x.A=1024\)
\(\Leftrightarrow A=2^9-\left(2^8+2^7+........+2+1\right)\)
\(\Leftrightarrow A=2^9-\left(2^9-1\right)\)
\(\Leftrightarrow A=1\)
\(\Leftrightarrow2^x.1=1024\)
\(\Leftrightarrow2^x=2^{10}\)
\(\Leftrightarrow x=10\)
Vậy ...
còn 1 câu nữa bn!!