a) Vì \(-|x+0,75|\le0;\forall x\)

\(5-|x+0,75|\le5-0;\forall x\)

Hay \(A\le5;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow|x+0,75|=0\)

                        \(\Leftrightarrow x=-0,75\)

Vậy MAX A=5 \(\Leftrightarrow x=-0,75\)

( tương tự ko giải đc ib )

A = 5 - |x + 0,75|

Ta có: |x + 0,75| \(\ge\)0 \(\forall\)x

=> 5 - |x + 0,75| \(\le\)5 \(\forall\)x

Dấu "=" xảy ra khi : x + 0,75 = 0 <=> x = -0,75

Vậy Max của A = 5 tại x = -0,75

B = |x - 5,2| + 6,5

Ta có: |x - 5,2| \(\ge\)0 \(\forall\)x

=> |x - 5,| + 6,5 \(\ge\)6,5 \(\forall\)x

Dấu "=" xảy ra khi: x - 5 = 0 <=> x = 5

Vậy Min của B = 6,5 tại x = 5

C = -|x + 4/5| - 6

Ta  có: -|x + 4/5| \(\le\)0 \(\forall\)x

=> -|x+ 4/5| - 6 \(\le\)-6 \(\forall\)x

Dấu "=" xảy ra khi: x + 4/5 = 0 <=> x = -4/5

Vậy Max của C = -6 tại x = -4/5

a) (2x - 3)² = 16

=> (2x - 3)² = 4²

=> \(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)

=> \(\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)

Vậy ...

b) 2.3^{x + 2} + 4. 3^{x + 1} = 10.3⁶

=> 2.3^{x + 1}. 3 + 4.3^{x + 1} = 10 . 3⁶

=> 6.3^{x + 1} + 4.3^{x + 1} = 10.3⁶

=> (6 + 4).3^{x + 1}= 10.3⁶

=> 10.3^{x + 1}= 10.3⁶

=> 3^{x + 1}= 3⁶

=> x + 1 = 6

=> x = 6 - 1

=> x = 5

\(a,\left(2x-3\right)^2=16\)

\(\Rightarrow\left(2x-3\right)^2=4^2\)

\(\Rightarrow2x-3=4\)

\(2x=4+3\)

\(2x=7\)

\(x=7:2\)

\(x=\frac{7}{2}\)

Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)

Ta có: \(\frac{8^{23}+1}{8^{24}+1}< \frac{8^{23}+1-1+8^5}{8^{24}+1-1+8^5}=\frac{8^{23}+8^5}{8^{24}+8^5}=\frac{8^5.\left(8^{18}+1\right)}{8^5.\left(8^{19}+1\right)}=\frac{8^{18}+1}{8^{19}+1}\)

Vậy \(\frac{8^{23}+1}{8^{24}+1}< \frac{8^{18}+1}{8^{19}+1}\)

Tham khảo lời giải tại link : https://olm.vn/hoi-dap/detail/90330086488.html

a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(\Rightarrow A+2A=2^{101}-2\)

  \(A\left(1+2\right)=2^{101}-2\)

  \(A.3=2^{101}-2\)

  \(A=\frac{2^{101}-2}{3}\)

b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)

\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)

\(\Rightarrow B+3B=3^{101}-3\)

\(B\left(1+3\right)=3^{101}-3\)

\(4B=3^{101}-3\)

   \(B=\frac{3^{101}-3}{4}\)

Thanks bạn

3^x + 1 = 9

3^x = 9 - 1

3^x = 8

-> x thuộc rỗng 

ĐÚNG CHO MK NHÉ!

3^x+1 = 3²

<=> x + 1 = 2

<=> x = 1

\(5^{\left(x-2\right).\left(x+3\right)}=1\)

mà \(5^0=1\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}\)

 \(5^{\left(x-2\right).\left(x+3\right)}\)=1

<=> (x-2)(x+3)=0

<=> +) x-2=0         +)x+3=0

          =>x=2             =>x=-3

Ta có : \(\left(x-2\right)^8=\left(x-2\right)^6=>\left(x-2\right)^8-\left(x-2\right)^6=0=>\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)

\(=>\orbr{\begin{cases}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{cases}}\)

\(=>\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)

\(=>\orbr{\begin{cases}x=2\\x-2=1 hoac x-2=-1\end{cases}}=>\orbr{\begin{cases}x=2\\x=3 hoac x=1\end{cases}}\)\(=>x=2\)hoăc \(x=3\) hoăc \(x=1\)

\(\left(x-2\right)^8=\left(x-2\right)^6\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^6=0\\\left(x-2\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=\pm1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3,x=1\end{cases}}}\)

_Tần vũ_

\(\frac{x}{y}=\frac{1,2}{2,5}\Rightarrow\frac{x}{y}=\frac{12}{25}\Rightarrow\frac{x}{12}=\frac{y}{25}\)

\(\Rightarrow\frac{x}{12}=\frac{y}{25}=\frac{y-x}{25-12}=\frac{26}{13}=2\)

\(\Rightarrow\hept{\begin{cases}x=2\cdot12=24\\y=2\cdot25=50\end{cases}}\)

vậy_

#)Giải :

Ta có : \(\frac{x}{y}=\frac{1,2}{2,5}\Rightarrow2,5x=1,2y\Rightarrow\frac{x}{1,2}=\frac{y}{2,5}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{y}{2,5}=\frac{x}{1,2}=\frac{y-x}{2,5-1,2}=\frac{26}{1,3}=20\)

\(\hept{\begin{cases}\frac{x}{1,2}=20\\\frac{y}{2,5}=20\end{cases}\Rightarrow\hept{\begin{cases}x=24\\y=50\end{cases}}}\)

Vậy x = 24; y = 50

Vì tam giác ABC cân tại A nên:

\(\widehat{ABC}=\widehat{ACB}=75^o\)

Áp dụng tính chất tổng ba góc trong tam giác HBC ta có:

\(\widehat{BHC}+\widehat{ACB}+\widehat{HBC}=180^o\)

\(90^o+75^0+\widehat{HBC}=180^o\)

\(165^o+\widehat{HBC}=180^o\)

\(\widehat{HBC}=180^o-165^o=15^o\)

Ta lại có: \(\widehat{ABC}=\widehat{HBA}+\widehat{HBC}\)

\(\Rightarrow\widehat{HBA}=\widehat{ABC}-\widehat{HBC}=75^o-15^o=60^o\)

Mặt khác: \(15^o=\frac{1}{4}60^o\)

Vậy nên \(\widehat{HBC}=\frac{1}{4}\widehat{HBA}\)