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4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)
=>3^x*23=3^16*(-107)
=>\(x\in\varnothing\)
2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)
=>2^x=2^10
=>x=10
3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>8^x=8^9
=>x=9
1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)
=>3^x=3^4*3^2
=>x=4+2=6
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
a) 4x2(5x2 + 3) – 6x(3x3 – 2x + 1) – 5x3 (2x – 1)
= 4x2 . 5x2 + 4x2 . 3 – [6x . 3x3 + 6x . (-2x) + 6x . 1] – [5x3 . 2x + 5x3 . (-1)]
= 20x4 + 12x2 – (18x4 – 12x2 + 6x) – (10x4 – 5x3)
= 20x4 + 12x2 - 18x4 + 12x2 - 6x - 10x4 + 5x3
= (20x4 – 18x4 - 10x4 ) + 5x3 + (12x2 + 12x2 ) – 6x
= -8x4 + 5x3 + 24x2 – 6x
\(\begin{array}{l}b)\dfrac{3}{2}x\left( {{x^2} - \dfrac{2}{3}x + 2} \right) - \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\\ = \dfrac{3}{2}x.{x^2} + \dfrac{3}{2}x.( - \dfrac{2}{3}x) + \dfrac{3}{2}x.2 - (\dfrac{5}{3}{x^2}.x + \dfrac{5}{3}{x^2}.\dfrac{6}{5})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - (\dfrac{5}{3}{x^3} + 2{x^2})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - \dfrac{5}{3}{x^3} - 2{x^2}\\ = (\dfrac{3}{2}{x^3} - \dfrac{5}{3}{x^3}) + ( - {x^2} - 2{x^2}) + 3x\\ = \dfrac{{ - 1}}{6}{x^3} - 3{x^2} + 3x\end{array}\)
a) (x-1):2/3=-2/5
=>x-1=-4/15
=>x=11/15
b) |x-1/2|-1/3=0
=>|x-1/2|=1/3
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\)
c) Tương Tự câu B
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)
⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)
⇒\(9+3x=28\)
⇒\(3x=19\)
⇒\(x=\dfrac{19}{3}\)
bạn thay vào là tìm được y
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9