\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)

Vậy \(x = \dfrac{{ - 9}}{2}\)

\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)

Vậy \(x = \dfrac{{ - 20}}{3}\)

Bài 1: 

Ta có: \(3x=2y\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=-15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)

Vậy: (x,y)=(-6;-9)

Bài 2: 

a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x+y-z=20

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)

Vậy: (x,y,z)=(40;30;50)

Bài 2: 

b) Ta có: \(\dfrac{y}{3}=\dfrac{z}{7}\)

nên \(\dfrac{y}{12}=\dfrac{z}{28}\)

mà \(\dfrac{x}{11}=\dfrac{y}{12}\)

nên \(\dfrac{x}{11}=\dfrac{y}{12}=\dfrac{z}{28}\)

hay \(\dfrac{2x}{22}=\dfrac{y}{12}=\dfrac{z}{28}\)

mà 2x-y+z=152

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{22}=\dfrac{y}{12}=\dfrac{z}{28}=\dfrac{2x-y+z}{22-12+28}=\dfrac{152}{38}=4\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{11}=4\\\dfrac{y}{12}=4\\\dfrac{z}{28}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=44\\y=48\\z=112\end{matrix}\right.\)

Vậy: (x,y,z)=(44;48;112)

Cái này bạn áp dụng tính chất 1 của tỉ lệ thức là ra ngay mà!

Hai tỉ số bằng nhau khi tích 2 ngoại tỉ bằng tích 2 trung tỉ.

a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(\Leftrightarrow7x-21=5x+25\)

\(\Leftrightarrow7x-5x=25+21\)

\(\Leftrightarrow2x=46\)

\(\Rightarrow x=\dfrac{46}{2}=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow7.9=\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow63=\left(x+1\right)x-\left(x+1\right)\)

\(\Leftrightarrow63=x^2+x-x-1\)

\(\Leftrightarrow63=x^2-1\)

\(\Leftrightarrow63+1=x^2\)

\(\Rightarrow64=x^2\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

\(\Leftrightarrow\left(x+4\right)^2=5.20\)

\(\Leftrightarrow\left(x+4\right)^2=100=10^2=\left(-10\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left(x-1\right)x+\left(x-1\right).3=\left(x-2\right)x+\left(x-2\right).2\)

\(\Leftrightarrow x^2-x+3x-3=x^2-2x+2x-4\)

\(\Leftrightarrow x^2+2x-3=x^2-4\)

\(\Leftrightarrow x^2+2x-3+4-x^2=0\)

\(\Leftrightarrow2x-3+4=0\)

\(\Leftrightarrow2x=-1\)

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

Chẳng biết có đúng không...

Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Leftrightarrow7x-21=5x+25\)

\(\Leftrightarrow7x-5x=21+25\)

\(\Leftrightarrow2x=46\)

\(\Rightarrow x=46:2=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Rightarrow x^2=\left(\pm8\right)^2\)

\(\Rightarrow x=8\) hoặc \(x=-8\)

2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(7x-21=5x+25\)

\(7x-5x+25=21\)

\(2x+25=21\)

\(2x=-4\Rightarrow x=-2\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(7.9=\left(x+1\right)\left(x-1\right)\)

\(63=x\left(x-1\right)+1\left(x-1\right)\)

\(63=x^2-x+x-1\)

\(x^2=63+1=64\)

\(x=\left\{\pm8\right\}\)

c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)

\(x\left(x+4\right)+4\left(x+4\right)=40\)

\(x^2+4x+4x+16=40\)

\(x^2+8x=40-16=24\)

\(x\left(x+8\right)=24\)

\(x\in\left\{\varnothing\right\}\)

d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)

\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)

\(x^2-2x+2x-4=x^2+3x-x-3\)

\(\)\(x^2-4=x^2+2x-3\)

\(\Leftrightarrow x^2-x^2-2x+3=4\)

\(-2x+3=4\)

\(-2x=1\)

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

bài 1) a) ta có : \(\dfrac{x}{10}=\dfrac{y}{15}\) \(\Leftrightarrow\) \(10y=15x\) \(\Leftrightarrow\) \(10y=15.\dfrac{7}{2}=\dfrac{105}{2}\)

\(\Leftrightarrow y=\dfrac{105}{2}:10=\dfrac{21}{4}\)

vậy \(x+2y-3z=20\Leftrightarrow\) \(\dfrac{7}{2}+2.\dfrac{21}{4}-3z=20\)

\(\Leftrightarrow\) \(-3z=20-14=6\) \(\Leftrightarrow\) \(z=-2\)

vậy \(x=\dfrac{7}{2}\) ; \(y=\dfrac{21}{4}\) ; \(z=-2\)

2.

a) \(\dfrac{x-1}{4}=\dfrac{9}{x-1}\)

\(\Leftrightarrow\left(x-1\right)^2=9.4\)

\(\Leftrightarrow\left(x-1\right)^2=36\)

\(\Leftrightarrow\left(x-1\right)^2=\left[{}\begin{matrix}6^2\\-6^2\end{matrix}\right.\)

\(\Leftrightarrow x-1=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\)

\(\Leftrightarrow x=\left[{}\begin{matrix}7\\-5\end{matrix}\right.\)

1. a. \(\dfrac{3}{-1}=\dfrac{-15}{5}\); \(\dfrac{-1}{3}=\dfrac{5}{-15}\)

\(\dfrac{3}{-15}\)= \(\dfrac{5}{-1}\); \(\dfrac{-15}{3}=\dfrac{-1}{5}\)

b. \(\dfrac{4}{-3}=\dfrac{-12}{9};\dfrac{-3}{4}=\dfrac{9}{-12}\)

\(\dfrac{4}{-12}=\dfrac{9}{-3};\dfrac{-12}{4}=\dfrac{-3}{9}\)

c. \(\dfrac{3}{7}=\dfrac{c}{b};\dfrac{7}{3}=\dfrac{b}{c}\)

\(\dfrac{3}{c}=\dfrac{b}{7};\dfrac{c}{3}=\dfrac{7}{b}\)

d. \(\dfrac{a}{b}=\dfrac{y}{x};\dfrac{b}{a}=\dfrac{x}{y}\)

\(\dfrac{a}{y}=\dfrac{x}{b};\dfrac{y}{a}=\dfrac{b}{x}\)

chúc bạn học tốt

Câu 2: 

a: =>(x-1)²=36

=>x-1=6 hoặc x-1=-6

=>x=7 hoặc x=-5

b: =>(x+2)²=100

=>x+2=10 hoặc x+2=-10

=>x=8 hoặc x=-12

a)\(\dfrac{x}{7}=\dfrac{18}{14}\)

\(\Rightarrow x=\dfrac{7.18}{14}=9\)

b)\(6:x=1\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{6}{x}=\dfrac{7}{4}\)

\(\Rightarrow x=\dfrac{6.4}{7}=\dfrac{24}{7}\)

c)5,7:0,35=(-x):0,45

\(\Leftrightarrow\dfrac{114}{7}=\dfrac{-x}{0,45}\)

\(\Rightarrow\left(-x\right)=\dfrac{114.0,45}{7}=\dfrac{-513}{70}\)

c,M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) :  \(\dfrac{\sqrt{x}+3}{\sqrt{x}+5}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) \(\times\) \(\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\) = \(\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}\)

 M = 1  - \(\dfrac{7}{\sqrt{x}+3}\) 

 M \(\in\) Z ⇔ 7 ⋮ \(\sqrt{x}\) + 3 vì \(\sqrt{x}\) ≥ 0 ⇒ \(\sqrt{x}\) + 3 ≥ 3 ⇒ 0< \(\dfrac{7}{\sqrt{x}+3}\) ≤ \(\dfrac{7}{3}\)

⇒ M Đạt giá trị nguyên lớn nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) đạt giá trị nguyên nhỏ nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) = 1 ⇔ \(\sqrt{x}\) + 3  = 7 ⇔ \(\sqrt{x}\) = 4 ⇔ \(x\) = 16 

Mnguyên(max)  = 1 - 1 = 0 xảy ra khi \(x\) = 16

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

\(\Leftrightarrow x=-\dfrac{3}{4}\)

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=-\dfrac{2}{3}\)

c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

#YVA6

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=\dfrac{-2}{3}\)

\(c,\dfrac{-1}{4}.x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)

\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

\(#yH\)

\(#NBaoNgoc\)