1.
a) \(\frac{3}{7}+\frac{-5}{2}+\frac{-3}{5}\\ =\frac{30}{70}+\frac{-175}{70}+\frac{-42}{70}\\ =\frac{30-175-42}{70}\\ =\frac{-187}{70}\)
b) \(\frac{-8}{18}-\frac{15}{27}\\ =\frac{-4}{9}-\frac{5}{9}\\ =\frac{-9}{9}=-1\)
c) \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\\ =\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\\ =\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\\ =\frac{56+20-49}{70}\\ =\frac{27}{70}\)
2.
a) \(x+\frac{1}{4}=\frac{4}{3}\\ x=\frac{4}{3}-\frac{1}{4}\\ x=\frac{16}{12}-\frac{3}{12}\\ x=\frac{13}{12}\)
Vậy \(x=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=\frac{-6}{7}\\ -x=\frac{-6}{7}+\frac{2}{3}\\ -x=\frac{-18}{21}+\frac{14}{21}\\ -x=-\frac{4}{21}\\ x=\frac{4}{21}\)
Vậy \(x=\frac{4}{21}\)
c) \(x^2=16\\ x^2=4^2=\left(-4\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{4;-4\right\}\)
d) \(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\\ \frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\\ \frac{5}{3}x=\frac{15}{21}-\frac{14}{21}\\ \frac{5}{3}x=\frac{1}{21}\\ x=\frac{1}{21}:\frac{5}{3}\\ x=\frac{1}{21}\cdot\frac{3}{5}\\ x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}\)
3.
a) Xét △AKB và △AKC có:
AB = AC
KB = KC
AK: cạnh chung
\(\Rightarrow\text{△AKB = △AKC (c.c.c) }\)
b) \(\text{△AKB = △AKC }\)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà \(\widehat{AKB}+\widehat{AKC}=180^o\) (2 góc kề bù)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}=90^o\\ \Rightarrow AK\perp BC\)
Câu 3:
a/ Xét ΔAKB và ΔAKC có:
AB = AC (GT)
\(\widehat{BAK}=\widehat{CAK}\left(GT\right)\)
AK: cạnh chung
=> ΔAKB = ΔAKC (c.g.c)
b/ VìΔAKB = ΔAKC (câu a)
\(\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà 2 góc này lại là hai góc kề bù
=> \(\widehat{AKB}=\widehat{AKC}=180^0:2=90^0\)
=> AK ⊥BC
Cau 2:
a) \(x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=-\frac{6}{7}\)
=> \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)
=> \(x=\frac{4}{21}\)
c) x2 = 16
=> x = 4 hoặc x =-4
