`#1231.2021`

`1.`

Ta có:

`y` tỉ lệ nghịch với `x` theo hệ số tỉ lệ `-4`

`=> y = (-4)/x` `(1)`

`x` tỉ lệ nghịch với `z` theo hệ số tỉ lệ `3/4`

`=> x = 3/4 \div z` `(2)`

Thay `(2)` vào `(1)`

`=> y = (-4)/(3/4 \div z) => y = -16/3 * z`

Vậy, `y` và `z` tỉ lệ thuận với nhau theo hệ số tỉ lệ `-16/3`

`=> A.`

Chọn D

a) \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{21}{21}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\dfrac{20}{21}\)

\(A=\dfrac{10}{21}\)

b) \(B=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{96.97}+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{96}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\dfrac{98}{99}\)

\(B=-\dfrac{97}{99}\)

\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)

c: -8/13+(-7/5-x)=-1/2

=>x+7/5+8/13=1/2

=>x=1/2-7/5-8/13=-197/130

d: \(\Leftrightarrow-\dfrac{8}{7}+\dfrac{5}{3}-\left(x-\dfrac{7}{3}\right)=\dfrac{-4}{21}\)

=>\(x-\dfrac{7}{3}=\dfrac{-8}{7}+\dfrac{5}{3}+\dfrac{4}{21}=\dfrac{-24+35+4}{21}=\dfrac{18}{21}=\dfrac{6}{7}\)

=>x=6/7+7/3=18/21+49/21=67/21

e: =>x:1/2=-2/3-2/5=-16/15

=>x=-16/15*1/2=-8/15

f: =>-8/5*x=-1/3+4/9=1/9

=>x=-1/9:8/5=-1/9*5/8=-5/72

g: =>-4/5x-1/4+x=-13/3

=>1/5x=-13/3+1/4=-52/12+3/12=-49/12

=>x=-49/12*5=-245/12

h: =>12/7:x-1/2=0 hoặc 2/5x-3/2=0

=>12/7:x=1/2 hoặc 2/5x=3/2

=>x=12/7:1/2=24/7 hoặc x=3/2:2/5=3/2*5/2=15/4

a: k=3/8

b: y=3/8x

Còn câu c thì sao ạ

Chọn A

80

(a) \(A=\dfrac{3}{x-2}\in Z\)

\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;0;2;4\right\}.\)

(b) \(B=-\dfrac{11}{2x-3}\in Z\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;1;2;7\right\}.\)

(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)

\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

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

Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)

(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)

\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)

câu (a) thiếu điều kiện x khác 2 rồi bạn êi

a: \(\widehat{C}=30^0\)

14:

a) \(\left(3x-1\right)^4=81\)

\(\Rightarrow\left(3x-1\right)^4=\left(\pm3\right)^4\)

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

\(\Rightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\)

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

b) \(\left(x+1\right)^5=-32\)

\(\Rightarrow\left(x+1\right)^5=\left(-2\right)^5\)

\(\Rightarrow x+1=-2\)

\(\Rightarrow x=-3\)

c) \(\left(x+\dfrac{2022}{2023}\right)^3=0\)

\(\Rightarrow x+\dfrac{2022}{2023}=0\)

\(\Rightarrow x=-\dfrac{2022}{2023}\)

d) \(\left(x-\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=\dfrac{3}{5}\\x-\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

e) \(\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{9}{10}\)

f) \(2^x+2^{x+3}=144\)

\(\Rightarrow2^x\cdot\left(1+2^3\right)=144\)

\(\Rightarrow2^x\cdot9=144\)

\(\Rightarrow2^x=16\)

\(\Rightarrow2^x=2^4\)

\(\Rightarrow x=4\)

16:

a: \(\Leftrightarrow2^{x+1}\cdot5^y=\left(2^2\cdot5\right)^x=2^{2x}\cdot5^x\)

=>2x=x+1 và x=y

=>x=1=y

b: \(\Leftrightarrow5^x\cdot3^{x-y}=3^y\cdot5^{2y}\)

=>x=2y và x-y=y

=>x=2y(luôn đúng)

Vậy: \(\left(x,y\right)\in N\)

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