2.(x-5)-3(x+7)=14
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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:
\(\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}=\dfrac{x+2y-3z}{3+8-18}=\dfrac{-14}{-7}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\\z=12\end{matrix}\right.\)
\(x\times\dfrac{3}{5}=\dfrac{2}{3}+\dfrac{1}{3}\)
\(x\times\dfrac{3}{5}=\dfrac{3}{3}\)
\(x\times\dfrac{3}{5}=1\)
\(x=1:\dfrac{3}{5}\)
\(x=1\times\dfrac{5}{3}\)
\(x=\dfrac{5}{3}\)
\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)
\(x-\dfrac{4}{9}=\dfrac{3}{7}\times\dfrac{14}{9}\)
\(x-\dfrac{4}{9}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{4}{9}\)
`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.
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: \(=\dfrac{7\cdot28\cdot15}{9\cdot25\cdot14}=\dfrac{1}{2}\cdot\dfrac{28}{25}\cdot\dfrac{5}{3}=\dfrac{140}{150}=\dfrac{14}{15}\)
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`
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
=> 2x-10-3x-21=14 <=> -x=45 <=> x=-45
2.(x - 5) - 3.(x + 7) = 14
=> 2x - 10 - 3x - 21 = 14
=> -x - 31 = 14
=> -x = 45
=> x = -45