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a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
1) 5.(3-x)+2.(x-7)=-14
15-5x+2x-14=-14
1-3x=-14
3x=15
X=5
2) 30.(x+2)-6.(x-5)-24x=100
30x+60-6x+30-24x=100
0X+90=100
0X=10 vô lí
=> ko có giá trị x thỏa mãn điều kiện
3) (3x-9)^2=36
3x-9=6
3x-9=-6
TH1:3x-9=6 TH2:3x-9=-6
3x=15 3x=3
X=5 x=1
Vậy….
4) (1-2x)^3=-27
(1-2x)3=(-3)3
1-2x=-3
2x=4
X=2
Vậy…
5) (x-3).(x-2)<0
=>x-3 và x-2 cùng dấu
TH1:x-3>0 TH2:x-3<0
x-2<0 x-2>0
=>X>3 =>x<3
X<2 x>2
=>x>3 =>x<2
Vậy 3<x<2
câu 6 chịuuuu
câu 5 hơi khó ko bt có đúng hay ko đâu :)))
3) \(\left(3x-9\right)^2=36\Leftrightarrow\orbr{\begin{cases}3x-9=6\\3x-9=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=15\\3x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)
4) \(\left(1-2x\right)^3=-27\)
<=> 1-2x=-3
<=> 3x=4
<=> \(x=\frac{4}{3}\)
5) (x-3)(x-2)<0
=> x-3 và x-2 trái dấu nhau
thấy x-3<x-2 => \(\hept{\begin{cases}x-3< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}\Leftrightarrow}2< x< 3}\)
6) làm tương tự
1: =>5(2x+6)=40
=>2x+6=8
=>2x=2
=>x=1
2: =>12-(x+3)=256:64=4
=>(x+3)=8
=>x=5
3: =>2x-1=3 hoặc 2x-1=-3
=>x=2 hoặc x=-1
4: \(\Leftrightarrow3^{x+2017}=3^{2015}\)
=>x+2017=2015
=>x=-2
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6