Ủa alo

Tui làm rồi mà

cái này bạn gửi ở dưới rồi còn gì

ABC cân vì có AB = AC

OAD đều vì nó là tổng 3 tam giác cân cộng lại.

a: \(3\dfrac{3}{7}:1\dfrac{5}{7}\)

\(=\dfrac{24}{7}:\dfrac{12}{7}\)

\(=\dfrac{24}{7}\cdot\dfrac{7}{12}=\dfrac{24}{12}=2\)

b: \(\dfrac{2}{3}+\dfrac{-3}{5}=\dfrac{2}{3}-\dfrac{3}{5}\)

\(=\dfrac{10-9}{15}\)

\(=\dfrac{1}{15}\)

c: \(\dfrac{2}{9}-\left(\dfrac{1}{20}+\dfrac{2}{9}\right)\)

\(=\dfrac{2}{9}-\dfrac{1}{20}-\dfrac{2}{9}\)

\(=-\dfrac{1}{20}\)

d: \(\dfrac{11}{23}\cdot\dfrac{12}{17}+\dfrac{11}{23}\cdot\dfrac{5}{17}+\dfrac{12}{23}\)

\(=\dfrac{11}{23}\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\dfrac{12}{23}\)

\(=\dfrac{11}{23}+\dfrac{12}{23}=\dfrac{23}{23}=1\)

Ta có

 \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right).\left(a+c\right)\\ Cmtt:b^2+1=\left(b+a\right).\left(b+c\right)\\ c^2+1=\left(c+a\right).\left(c+b\right)\)

Nên

 \(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\\ =\dfrac{\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{\left(c-a\right)}{\left(b+c\right)\left(b+a\right)}+\dfrac{\left(a-b\right)}{\left(c+a\right)\left(c+b\right)}\\ =\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =0\)

\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\)

\(=\dfrac{b-c}{a^2+ab+bc+ac}+\dfrac{c-a}{b^2+ab+bc+ca}+\dfrac{a-b}{c^2+ab+bc+ca}\)

\(=\dfrac{b-c}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{c-a}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{a-b}{c\left(c+a\right)+b\left(a+c\right)}\)

\(=\dfrac{b-c}{\left(a+c\right)\left(a+b\right)}+\dfrac{c-a}{\left(b+c\right)\left(a+b\right)}+\dfrac{a-b}{\left(b+c\right)\left(a+c\right)}\)

\(=\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(a+c\right)+\left(a-b\right)\left(a+b\right)}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)

\(=\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\) 

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2+9-6x}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

b) \(\dfrac{x+2}{3x}+\dfrac{x-5}{5x}+\dfrac{x+8}{4x}=\dfrac{20\left(x+2\right)+12\left(x-5\right)+15\left(x+8\right)}{60x}=\dfrac{47x+100}{60x}\)

gãy đốt sống cổ rùi :(((((

Câu 7: B

Câu 8: D

144

4 giờ 23 phút 

ko bít có đúng ko nữa

Chọn C