\(a,\Leftrightarrow m-2=3\Leftrightarrow m=5\\ b,y=f\left(x\right)=\left(5-2\right)x=3x\\ \Leftrightarrow f\left(3\right)+\dfrac{1}{3}f\left(-2\right)=9+\dfrac{1}{3}\cdot\left(-6\right)=7\)

Bài 5: 

\(A=2A-A=2^2+2^3+...+2^{107}-2-2^2-...-2^{2016}=2^{107}-2\)

\(2\left(A+2\right)=2^{2x}\\ \Rightarrow2\left(2^{107}-2+2\right)=2^{2x}\\ \Rightarrow2^{108}=2^{2x}\\ \Rightarrow2x=108\\ \Rightarrow x=54\)

Bài 3:

Gọi số học sinh lớp 7A, 7B lần lượt là a,b

Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{9}\\y-x=5\end{matrix}\right.\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{8}=\dfrac{y}{9}=\dfrac{y-x}{9-1}=\dfrac{5}{1}=5\)

\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{9}=5\Rightarrow y=45\)

Vậy số học sinh lớp 7A, 7B lần lượt là 40, 45 học sinh

Bài 9:

\(\widehat{B}=\widehat{HAC}=60^0\)

cách giải cụ thể luôn đc ko ạ

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)

5: \(=\dfrac{1}{2}\cdot10-\dfrac{1}{2}=\dfrac{1}{2}\cdot9=\dfrac{9}{2}\)

lx đề r bn

Câu hỏi đâu bạn?

10³ + 2¹⁵

= 1000 + 32768

= 33768

Mà 33768 : 33 = 1023 (dư 9)

Em xem lại đề

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