K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

21 tháng 6 2017

1)

a) \(0,25^x\cdot12^x=243\)

\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(38^y:19^y=512\)

\(\Leftrightarrow2y\cdot y=512\)

\(\Leftrightarrow2y^2=512\)

\(\Leftrightarrow y^2=256\)

\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)

Vậy \(y_1=-16;y_2=16\)

2)

a) \(3^x+3^{x+2}=2430\)

\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)

\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)

\(\Leftrightarrow10\cdot3^x=2430\)

\(\Leftrightarrow3^x=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(2^{x+3}-2^x=224\)

\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)

\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)

\(\Leftrightarrow7\cdot2^x=224\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

3)

a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)

b) \(\left(x+0,7\right)^3=-27\)

\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x+\dfrac{3}{10}=-3\)

\(\Leftrightarrow x=-3-\dfrac{3}{10}\)

\(\Leftrightarrow x=-\dfrac{37}{10}\)

Vậy \(x=-\dfrac{37}{10}\)

4)

a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)

b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)

\(\Leftrightarrow2x-1=1\)

\(\Leftrightarrow2x=1+1\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

21 tháng 6 2017

1. a) \(0,25^x.12^x=243\)

\(\Rightarrow\left(0,25.12\right)^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

b) \(38^y:19^y=512\)

\(\Rightarrow\left(38:19\right)^y=512\)

\(\Rightarrow2^y=2^9\)

\(\Rightarrow y=9\)

Vậy \(y=9.\)

2) a) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x\left(1+9\right)=2430\)

\(\Rightarrow3^x=243=3^5\)

\(\Rightarrow x=5\)

Vậy x=5.

b) \(2^{x+3}-2^x=224\)

\(\Rightarrow2^x\left(8-1\right)=224\)

\(\Rightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

Vậy x=5.

Bài 3: dễ tự làm.

21 tháng 6 2017

a)0,25x.12x=243

(0,25x12)x=243

3x=243

3x=35

=>x=5

26 tháng 10 2017

tìm các số tự nhiên x, y biết: (x+3).(y-5)=38

giúp tôi trả lời câu này với! 

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)

12 tháng 9 2021

\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)

21 tháng 6 2017

a) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x+3^x.3^2=2430\)

\(\Rightarrow3^x\left(1+9\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=243=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

b) \(2^{x+3}-2^x=224\)

\(\Rightarrow2^x.8-2^x=224\)

\(\Rightarrow2^x\left(8-1\right)=224\)

\(\Rightarrow2^x.7=224\)

\(\Rightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

29 tháng 12 2021

Bài 1:

\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)

Bài  2:

\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)

Bài 1: 

Ta có: \(3x=2y\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=-15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)

Vậy: (x,y)=(-6;-9)

Bài 2: 

a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x+y-z=20

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)

Vậy: (x,y,z)=(40;30;50)