Bài 7: Tỉ lệ thức

c: \(-\dfrac{1}{5}:\left(1\dfrac{2}{5}\right)=-\dfrac{2}{7}:x\)

=>\(-\dfrac{2}{7}:x=-\dfrac{1}{5}:\dfrac{7}{5}=\dfrac{-1}{5}\cdot\dfrac{5}{7}=-\dfrac{1}{7}\)

=>\(x=\dfrac{-2}{7}:\dfrac{-1}{7}=\dfrac{2}{7}\cdot\dfrac{7}{1}=2\)

d: \(1\dfrac{1}{2}:\dfrac{2}{3}=x:\dfrac{1}{9}\)

=>\(x:\dfrac{1}{9}=\dfrac{3}{2}\cdot\dfrac{3}{2}=\dfrac{9}{4}\)

=>\(x=\dfrac{9}{4}\cdot\dfrac{1}{9}=\dfrac{1}{4}\)

\(1,8:1,3=\left(-2,7\right):\left(5x\right)\\ \Leftrightarrow\dfrac{1,8}{1,3}=\dfrac{-2,7}{5x}\\ \Rightarrow1,8.5x=1,3.\left(-2,7\right)\\ \Leftrightarrow9x=-3,51\\ \Leftrightarrow x=-0,39\)

Áp dụng `a/b=c/d -> ad=bc.`

`=> D.`

Chọn ý trên cùng

Bài 2

Xét ∆ABC và ∆ABD có:

AC = AD (gt)

AB là cạnh chung

∠BAC = ∠BAD (gt)

⇒ ∆ABC = ∆ABD (c-g-c)

Bài 3:

a: Xét ΔMHK và ΔPHN có

HM=HP

\(\widehat{MHK}=\widehat{PHN}\)

HK=HN

Do đó: ΔMHK=ΔPHN

b: ΔMHK=ΔPHN

=>\(\widehat{HMK}=\widehat{HPN}\)

mà hai góc này là hai góc ở vị trí so le trong

nên MK//NP

Xét ΔHMN và ΔHPK có

HM=HP

\(\widehat{MHN}=\widehat{PHK}\)

HN=HK

Do đó: ΔHMN=ΔHPK

=>\(\widehat{HMN}=\widehat{HPK}\)

mà hai góc này là hai góc ở vị trí so le trong

nên MN//KP

\(2\cdot12=4\cdot6\)

=>\(\dfrac{2}{4}=\dfrac{6}{12};\dfrac{2}{6}=\dfrac{4}{12};\dfrac{4}{2}=\dfrac{12}{6};\dfrac{6}{2}=\dfrac{12}{4}\)

\(6\cdot8=12\cdot4\)

=>\(\dfrac{6}{4}=\dfrac{12}{8};\dfrac{6}{12}=\dfrac{4}{8};\dfrac{4}{6}=\dfrac{8}{12};\dfrac{12}{6}=\dfrac{8}{4}\)

\(9\cdot8=12\cdot6\)

=>\(\dfrac{9}{12}=\dfrac{6}{8};\dfrac{9}{6}=\dfrac{12}{8};\dfrac{12}{9}=\dfrac{8}{6};\dfrac{6}{9}=\dfrac{8}{12}\)

`#3107.101107`

`(2x - 3)^2 = 36`

`=> (2x - 3)^2 = (+-6)^2`

`=>`\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `x \in {-3/2; 9/2}.`

\(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy: ...

Lời giải:

$\frac{ab}{bc}=\frac{b}{c}\Rightarrow \frac{a}{c}=\frac{b}{c}\Rightarrow a=b$

Cho $a=b=1, c=2$ thì:

$\frac{a^2+b^2}{b^2+c^2}=\frac{1^2+1^2}{1^2+2^2}=\frac{2}{5}$

$\frac{a}{c}=\frac{1}{2}$

Vì $\frac{2}{5}\neq \frac{1}{2}$ nên đề sai.

\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)

\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)

\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)