\(3^x=\frac{9^{10}}{27^5}=243\)\(=3^5\)\(\Rightarrow x=5\)

\(x+\frac{2}{15}=\frac{1}{3}\)

\(x=\frac{1}{3}-\frac{2}{15}\)

\(x=\frac{1}{5}\)

h, \(h,\frac{1}{3}-\frac{2}{3}:x=\frac{1}{4}\)

\(\frac{2}{3}:x\)= \(\frac{1}{3}-\frac{1}{4}\)

\(\frac{2}{3}:x=\frac{1}{12}\)

\(x=\frac{2}{3}:\frac{1}{12}\)

\(x=8\)

\(A=\left(x^2+3x+4\right)^2\)

ta có:

\(x^2+3x+4=x^2+2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\\ =\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

vậy \(minA=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\Leftrightarrow x=-\dfrac{3}{2}\)

a) ( 2011 - 125 x 8 : 4 ) x ( ( 48 : 8 x 6 )

= ( 2011 - 1000 : 4 ) x ( 6 x 6 )

= ( 2011 - 250 ) x 36

= 1761 x 36

= 63396

b) B = Không hiểu

c) C = 

\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|+\left|x-5\right|\)

\(\Rightarrow A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|4-x\right|+\left|5-x\right|\)

\(\Rightarrow A\ge x-1+x-2+0+4-x+5-x\)

\(\Rightarrow A\ge6\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-1\ge0;x-2\ge0\\x-3=0\\x-4\le0;x-5\le0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ge2\\x=3\\x\le4\end{cases}}\Rightarrow x\in\left(2;3;4\right)\)

Vậy MinA = 6 \(\Leftrightarrow x\in\left(2,3,4\right)\)

Phạm Tuấn Đạt dùng lý thuyết nào vậy?

Hơi tắt quá, chi tiết 1 chút đc ko ?

( 15 - x ) + ( x - 12 ) = ( -5 + x )

( 15 - 12 ) - x + x = -5 + x

3 = -5 + x

x = 3 + 5

x = 8

(x+1)+(x+3)+(x+5)+...+(x+99)=0

( x + x + x + .... + x ) + ( 1 + 3 + 5 + ... + 99 ) = 0

50x + 2500 = 0

50x = -2500

x = -2500 : 50

x = -50

trả lời hết lun đi