\(4^{x+2}-5\cdot4^x=176\)

\(\Rightarrow4^x\cdot4^2-5\cdot4^x=176\)

\(\Rightarrow4^x\cdot\left(16-5\right)=176\)

\(\Rightarrow4^x\cdot11=176\)

\(\Rightarrow4^x=\dfrac{176}{11}\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

\(4^{x+2}-5.4^x=176\)

\(=>4^x.4^2-5.4^x=176\)

\(=>4^x.\left(4^2-5\right)=176\)

\(=>4^x.\left(16-5\right)=176\)

\(=>4^x.11=176\)

\(=>4^x=176:11\)

\(=>4^x=16\)

\(=>4^x=4^2\)

\(=>x=2\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

\(\dfrac{2x-3}{5}=\dfrac{x-2}{4}\)
\(\Rightarrow\left(2x-3\right)\cdot4=\left(x-2\right)\cdot5\)
\(\Rightarrow8x-12=5x-10\)
\(\Rightarrow8x-5x=-10+12\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)
#kễnh

\(\dfrac{2x-3}{5}=\dfrac{x-2}{4}\)

\(\Leftrightarrow4\left(2x-3\right)=5\left(x-2\right)\)

\(\Leftrightarrow8x-12=5x-10\)

\(\Leftrightarrow3x=2\)

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

\(\frac{2}{7}\cdot\frac{x}{3}\cdot\frac{4}{5}=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\)

\(\frac{2}{7}\cdot\frac{x}{3}\cdot\frac{4}{5}=\frac{16}{35}\)

\(\Rightarrow\frac{16}{35}:\frac{4}{5}:\frac{2}{7}\)

\(\Rightarrow2\)

Mà : \(2=\frac{x}{3}\)

Suy ra số \(x\)cần tìm là:

\(2\cdot3=6\)

    Đáp số : \(6\)

Ai trả lời mình với

1. Tìm x là số tự nhiên

a) 5,32 < X x 2 > 6,01

TH1 : 1 x 2 = 2 ;         5,32 < 2 > 6,01 ⇒ Không thỏa mãn

TH2:  2 x 2 = 4  ;        5,32 < 4 > 6,01 ⇒ Không thỏa mãn

TH3: 3 x 2 = 6  ;         5,32 < 6 > 6,01 ⇒ Thỏa mãn

Vậy x = 2 ;  5,32 < 3 x 2 > 601

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

`a)4/5:3/x xx2/11=24/165`

`(4/5xx2/11)xx 3/x=8/55`

`8/55xx3/x=8/55`

`3/x=8/55:8/55`

`3/x=1`

`3:x=1`

`x=3:1`

`x=3`

____________________________________________

`b)4/3:x/5=5/13xx4/9`

`4/3:x/5=20/117`

`x/5=4/3:20/117`

`x/5=39/5`

`x=39`

A) 4/5:3/x X 2/11=24/165

\(\dfrac{4}{5}:\dfrac{3}{x}X\dfrac{2}{11}=\dfrac{24}{165}\)

\(\dfrac{3}{x}=\dfrac{24}{165}:\dfrac{4}{5}:\dfrac{2}{11}\)

\(\dfrac{3}{x}=1\)

\(\dfrac{3}{x}=\dfrac{3}{3}\)

=> x=3

B) \(\dfrac{4}{3}:\dfrac{x}{5}=\dfrac{5}{13}x\dfrac{4}{9}\)

\(\dfrac{4}{3}:\dfrac{x}{5}=\dfrac{20}{117}\)

\(\dfrac{x}{5}=\dfrac{4}{3}:\dfrac{20}{117}\)

\(\dfrac{x}{5}=\dfrac{39}{5}\)

=> x = 39

a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà -3<x<30

nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)

b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)

mà -16<=x<20

nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)

c: \(\Leftrightarrow x-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

d: \(\Leftrightarrow2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)

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