Bạn chụp lại đề nha bạn

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

1+5+3+9+5+7=30

Bằng 30 nha

k Đúng cho mình nhá

30 nhe bn

Bài 2: 

a: Vì (d) có hệ số góc là 3 nên a=3

Vậy: (d): y=3x+b

Thay x=1 và y=0 vào (d), ta được:

b+3=0

hay b=-3

b: Vì (d)//y=0,5x-2 nên a=0,5

Vậy: (d): y=0,5x+b

Thay x=0 và y=2 vào (d), ta được:

b=2

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

cần bài nào?

Chỉ thấy bài 5 với 6:

5.

\(f'\left(x\right)+2f\left(x\right)=0\Leftrightarrow f'\left(x\right)=-2f\left(x\right)\Leftrightarrow\dfrac{f'\left(x\right)}{f\left(x\right)}=-2\)

Lấy nguyên hàm 2 vế:

\(\int\dfrac{f'\left(x\right)}{f\left(x\right)}dx=\int-2dx\Rightarrow ln\left(f\left(x\right)\right)=-2x+C\)

Thay \(x=1\Rightarrow0=-2+C\Rightarrow C=2\)

\(\Rightarrow ln\left(f\left(x\right)\right)=-2x+2\Rightarrow f\left(x\right)=e^{-2x+2}\)

\(\Rightarrow f\left(-1\right)=e^4\)

6.

\(f\left(x\right)+x.f'\left(x\right)=2x+1\)

\(\Leftrightarrow x'.f\left(x\right)+x.f'\left(x\right)=2x+1\)

\(\Leftrightarrow\left[x.f\left(x\right)\right]'=2x+1\)

Lấy nguyên hàm 2 vế:

\(\int\left[x.f\left(x\right)\right]'dx=\int\left(2x+1\right)dx\)

\(\Rightarrow x.f\left(x\right)=x^2+x+C\)

Thay \(x=1\Rightarrow1.f\left(1\right)=1+1+C\Rightarrow C=1\)

\(\Rightarrow f\left(x\right)=\dfrac{x^2+x+1}{x}\)

\(\Rightarrow f\left(2\right)=\dfrac{7}{2}\)

Ủa sao đề khác rồi:

3.

\(f'\left(x\right)=-e^x.f^2\left(x\right)\Leftrightarrow\dfrac{f'\left(x\right)}{f^2\left(x\right)}=-e^x\)

Lấy nguyên hàm 2 vế:

\(\int\dfrac{f'\left(x\right)}{f^2\left(x\right)}dx=\int-e^xdx\)

\(\Rightarrow\dfrac{1}{f\left(x\right)}=e^x+C\)

Thay \(x=0\Rightarrow2=1+C\Rightarrow C=1\)

\(\Rightarrow\dfrac{1}{f\left(x\right)}=e^x+1\Rightarrow f\left(x\right)=\dfrac{1}{e^x+1}\)

\(\Rightarrow f\left(ln2\right)=\dfrac{1}{3}\)