Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}\)

Áp dung tcdtsbn , ta có: 

\(\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}=\dfrac{x+2y-3z}{3+8-18}=\dfrac{-14}{-7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\\z=12\end{matrix}\right.\)

áp dụng t/c dtsbn ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+2y-3z}{3+2.4-3.6}=\dfrac{-14}{-7}=2\)

\(\dfrac{x}{3}=2\Rightarrow x=6\\ \dfrac{y}{4}=2\Rightarrow y=8\\ \dfrac{z}{6}=2\Rightarrow z=12\)

\(x\times\dfrac{3}{5}=\dfrac{2}{3}+\dfrac{1}{3}\)

\(x\times\dfrac{3}{5}=\dfrac{3}{3}\)

\(x\times\dfrac{3}{5}=1\)

\(x=1:\dfrac{3}{5}\)

\(x=1\times\dfrac{5}{3}\)

\(x=\dfrac{5}{3}\)

\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)

\(x-\dfrac{4}{9}=\dfrac{3}{7}\times\dfrac{14}{9}\)

\(x-\dfrac{4}{9}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{4}{9}\)

`x`\(=\dfrac{18}{27}+\dfrac{12}{27}\)

\(x=\dfrac{30}{27}=\dfrac{10}{9}\)

Câu 1: Tác phẩm Quốc âm thi tập là của ai ? (1đ)

a. Nguyễn Trãi.

b. Lê Thánh Tông.

c. Lý Tử Tấn.

d .Lương Thế Vinh.

B1

A=11x^2-x-2

B=2(-4+x)

B2

a)=(x+3)^2(x-3)

B2

b)=(5x-7)(x+2)

Bài 1:

a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

Bài 2:

a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)

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

c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)

\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)

\(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{2\times8\times3}=2.\\ \dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{5\times2\times3\times5\times5\times7\times8}{3\times5\times5\times8\times2\times2\times3\times2\times7}=\dfrac{5}{12}.\)

a: \(=\dfrac{48-45}{54}\cdot\dfrac{9}{4}=\dfrac{3}{54}\cdot\dfrac{9}{4}=\dfrac{27}{216}=\dfrac{1}{8}\)

b: \(=\dfrac{6}{5}+1=\dfrac{11}{5}\)

c: \(=\dfrac{7\cdot28\cdot15}{9\cdot25\cdot14}=\dfrac{1}{2}\cdot\dfrac{28}{25}\cdot\dfrac{5}{3}=\dfrac{140}{150}=\dfrac{14}{15}\)

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

<=> 14 . x = 18 . 7

<=> 14x = 126

<=> x = \(\dfrac{126}{14}=9\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`2+(x+3)=7`

`\Rightarrow x+3=7-2`

`\Rightarrow x+3=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`5+(3+x)=10`

`\Rightarrow 3+x=10-5`

`\Rightarrow 3+x=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`(4+x)+1=7`

`\Rightarrow 4+x=7-1`

`\Rightarrow 4+x=6`

`\Rightarrow x=6-4`

`\Rightarrow x=2`

`(x+5)+3=9`

`\Rightarrow x+5=9-3`

`\Rightarrow x+5=6`

`\Rightarrow x=6-5`

`\Rightarrow x=1`

`(x-1)-4=7`

`\Rightarrow x-1=7+4`

`\Rightarrow x-1=11`

`\Rightarrow x=11+1`

`\Rightarrow x=12`

`4-(6-x)=1`

`\Rightarrow 6-x=4-1`

`\Rightarrow 6-x=3`

`\Rightarrow x=6-3`

`\Rightarrow x=3`

\(2+\left(x+3\right)=7\)

\(\Rightarrow2+x+3=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(5+\left(3+x\right)=10\)

\(\Rightarrow5+3+x=10\)

\(\Rightarrow x+8=10\)

\(\Rightarrow x=2\)

\(\left(4+x\right)+1=7\)

\(\Rightarrow4+x+1=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(\left(x+5\right)+3=9\)

\(=x+5+3=9\)

\(\Rightarrow x+8=9\)

\(\Rightarrow x=1\)

\(\left(x-1\right)-4=7\)

\(\Rightarrow x-1-4=7\)

\(\Rightarrow x-5=7\)

\(\Rightarrow x=12\)

\(4-\left(6-x\right)=1\)

\(\Rightarrow4-6-x=1\)

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

\(\Rightarrow x=-3\)