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Bài 4:
a: Xét ΔABD và ΔHBD có
BA=BH
\(\widehat{ABD}=\widehat{HBD}\)
BD chung
Do đó: ΔABD=ΔHBD
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\\ \Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{5}{8}\\ \Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\\ \Rightarrow x=\dfrac{16}{19}\)
e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\\ \Rightarrow6,5-2x=\dfrac{1}{2}\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\\ \Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)
g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\\ \Rightarrow5x-15=6x+9\\ \Rightarrow-x=24\\ \Rightarrow x=-24\)
h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\\ \Rightarrow\left(x-5\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x-5=-6\\x-5=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)
Bài 1:
c: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3k\\y=4k\end{matrix}\right.\)
Ta có: xy=48
\(\Leftrightarrow12k^2=48\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}y=3k=3\cdot2=6\\y=4k=4\cdot2=8\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}y=3k=3\cdot\left(-2\right)=-6\\y=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)
Bài 3:
a) Ta có: \(A-\left(9x^3+8x^2-2x-7\right)=-9x^3-8x^2+5x+11\)
\(\Leftrightarrow A=-9x^3-8x^2+5x+11+9x^3+8x^2-2x-7\)
\(\Leftrightarrow A=3x+4\)
b) Đặt A(x)=0
nên 3x+4=0
hay \(x=-\dfrac{4}{3}\)
Bạn có biết giải bài hình k giúp mình với 21:00 mình phải nộp rồi
4: Ta có: x+2x+...+18x=1368
\(\Leftrightarrow153x=1368\)
hay \(x=\dfrac{152}{17}\)
Ex1 ko có verb à
Ex2
1 like
2 listen
3 wears
4 teaches
5 do
6 goes
7 comes
8 goes
9 watch
10 walks
11 is
12 washes
13 studies
14 wants
15 plays
16 buys
17 studies
18 tries
19 washes
20 cries
21 says
22 flies
23 don't have
5.B
6.C
7.B