\(\dfrac{x-1}{2}=\dfrac{y}{3}=\dfrac{x-1+y}{2+3}=\dfrac{10}{5}=2\)
\(=>x-1=4=>x=5=>y=6\)

ta có:
x-1/2=y/3
áp dụng:
x-1/2=y/3=x-1+y/2+3=x+y-1/5=11-1/5=2
⇒x-1=4⇔x=3
y=3.2⇔y=6

a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/3 = y/4 = x/3 + y/4 = 28/7 = 4

=> x = 4 × 3 = 12

=> y = 4 × 4 = 16

Vậy x = 12, y = 16

B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1

=> x = -1 × 2 = -2

=> y = -1 × -5 = 5

Vậy x = -2, y = 5

C) làm tương tự như bài a, b

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

Câu đầu em xem lại đề bài sao có hai dấu bằng.

Câu 2: 

\(\dfrac{3}{2}\) \(\times\)y - \(\dfrac{3}{4}\) \(\times\)y + y = \(\dfrac{4}{5}\)

y \(\times\) ( \(\dfrac{3}{2}\) - \(\dfrac{3}{4}\) + 1) = \(\dfrac{4}{5}\)

y \(\times\) (\(\dfrac{6}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{4}{4}\)) = \(\dfrac{4}{5}\)

y \(\times\) \(\dfrac{7}{4}\)            = \(\dfrac{4}{5}\)

y = \(\dfrac{4}{5}\): \(\dfrac{7}{4}\)

y = \(\dfrac{16}{35}\)

a) Ta có : \(x - 2xy + y - 3 = 0\)

\(\Rightarrow-2xy+x+y=3\)

\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)

\(\Rightarrow4xy-2x-2y=-6\)

\(\Rightarrow4xy-2x-2y+1=-6+1\)

\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)

\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)

Tự lập bảng đi -.-

Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)

\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{11}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=11k\end{matrix}\right.\)\(\Rightarrow xyz=528k^3=-528\Rightarrow k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=8.\left(-1\right)=-8\\y=6.\left(-1\right)=-6\\z=11.\left(-1\right)=-11\end{matrix}\right.\)

7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)

Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)

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

⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

4) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)

\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)

\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)

\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)

6) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)

\(\dfrac{x+11}{13}=1\Rightarrow x=2\)

\(\dfrac{y+12}{13}=1\Rightarrow y=1\)

\(\dfrac{z+13}{15}=1\Rightarrow z=2\)

7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)

\(\Rightarrow x=4k,y=5k\)

\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)

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

\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)

b) \(\left|2x+3\right|=x+2\)

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

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

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

Đính chính

Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2}-\dfrac{x^2}{5}+\dfrac{y^2}{3}-\dfrac{y^2}{5}+\dfrac{z^2}{4}-\dfrac{z^2}{5}=0\)

\(\Leftrightarrow\dfrac{3}{10}x^2+\dfrac{2}{15}y^2+\dfrac{1}{20}z^2=0\)

\(\Leftrightarrow x=y=z=0\)

B1:
a.(-14/39)
b.45 độ
B2:
a.x=(-13/10)
b.
c.0

Bài 1:

a/\(\dfrac{17}{3}-\dfrac{5}{3}=\dfrac{17-5}{3}=\dfrac{12}{3}\)=4

b/Tam giác ABC có:

góc A+góc B+góc C=180 độ

=>70 độ+65 độ+góc C=180 độ

=>góc C =180 độ-70 độ-65 độ=45 độ

Bài 2:

a/\(x+3\dfrac{1}{2}=\dfrac{11}{5}=>x+\dfrac{7}{2}=\dfrac{11}{5}=>x=\dfrac{11}{5}-\dfrac{7}{2}=\dfrac{-13}{10}\)

c/\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right):\dfrac{4}{31}+\left(\dfrac{-4}{7}+\dfrac{6}{11}\right):\dfrac{4}{31}\)

=>\(\left(\dfrac{-3}{7}+\dfrac{5}{11}\right).\dfrac{31}{4}+\left(\dfrac{-4}{7}+\dfrac{6}{11}\right).\dfrac{31}{4}\)

=>\(\dfrac{31}{4}.\left(\dfrac{-3}{7}+\dfrac{5}{11}+\dfrac{\left(-4\right)}{7}+\dfrac{6}{11}\right)=\dfrac{31}{4}.0=0\)