Mình nhầm, để làm lại:

2^x + 2^{x + 3} = 144

<=> 2^x + 2^x . 2³ = 144

<=> 2^x(2³ + 1) = 144 

<=> 2^x . 9 = 144

<=> 2^x       = 144 : 9

<=> 2^x       = 16

<=> 16       = 2⁴

<=> x          = 4

Vậy x = 4

x = 4 , ủng hộ mk nha

2^x+2^x+3=144

=>2^x+2^x+3=2⁴+2⁷

=>2^x+2^x+3=2⁴+2⁴⁺³

=>x=4

\(\frac{x-1}{-3}\)=\(\frac{2x+5}{5}\)\(\Rightarrow\)5(x-1)=-3(2x+5)

                                           \(\rightarrow\)5x-5=-6x-15

                                          \(\rightarrow\)5x+6x=5-15

                                         \(\rightarrow\)11x=-10\(\Rightarrow\)x=\(\frac{-10}{11}\)

vậy.........................

ko hiểu

uk 276 nha mik tính là vậy còn ko biết đúng ko nữa cho mik 1 k nha hihi / HT/

Tl

Cho Đầu Bài Kiểu Gì Thế

?

HT

a,=(x\(^2\)-6x+9)+10-9

=(x-3)\(^2\)+1

Mà(x-3)\(^2\)\(\ge\)0

nên (x-3)\(^2\)+1>0

b,=  -(-4x+x\(^2\))-5

=    -(4-4x+x\(^2\))-5+4

=     -(2-x)\(^2\)-1

Mà  -(2-x)\(^2\)\(\le\)0

nên -(2-x)\(^2\)-1<   0

Võ Hoàng Tiên: Cảm ơn pạn nhiều lắm =)) nek :3 Hí Hí :)  Thankssssss 

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

Ta có: a+ b + c = 0

=> a+b = - c

a^3 + b^3 + c^3 = (a+b)³ - 3a²b - 3ab² + c³

                               = ( -c)³ - 3a²b - 3ab² + c³

                               = (-c)³ +c³ -3ab( a+b)

                       =   - 3ab (-c) = 3abc ( đpcm)

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}\) và \(x-3y=20\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{5}=\dfrac{3y}{9}=\dfrac{z}{2}=\dfrac{x-3y}{5-9}=\dfrac{20}{-4}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-5< =>x=-25\\\dfrac{y}{3}=-5< =>y=-15\\\dfrac{z}{2}=-5< =>z=-10\end{matrix}\right.\)

Vậy ....

\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\left(x+2\right)^2=\left(2x-1\right)^2\)

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

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

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

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

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

(x+2)2=(2x−1)2⇔(x+2)2−(2x−1)2=0⇔[x+2−(2x−1)][x+2+2x−1]=0⇔(x+2−2x+1)(x+2+2x−1)=0⇔(−x+3)(3x+1)=0⇔[−x+3=03x+1=0​⇔[−x=−33x=−1​
 

=) x = 3 hoặc -1/3.

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