a,A,C,D

b,C

a) Câu A;D

b) Câu C

mình hứa sẽ tick

= 2/5 + 1/5 + 1/5

= 4/5

b. = 2/7 + 4/7 + 5/21

= 6/21 + 12/21 + 5/21

= 23/21

HT

a: ĐKXĐ: a>=0; a<>4

\(H=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)

\(=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\frac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)

b: H<2

=>H-2<0

=>\(\frac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}<0\)

=>\(\frac{-\sqrt{a}}{\sqrt{a}-2}<0\)

=>\(\sqrt{a}-2>0\)

=>\(\sqrt{a}>2\)

=>a>4

c: \(a^2+3a=0\)

=>a(a+3)=0

=>a=0(nhận) hoặc a=-3(loại)

Thay a=0 vào H, ta được:

\(H=\frac{0-4}{0-2}=\frac{-4}{-2}=2\)

d: H=5

=>\(\sqrt{a}-4=5\left(\sqrt{a}-2\right)\)

=>\(5\sqrt{a}-10=\sqrt{a}-4\)

=>\(4\sqrt{a}=6\)

=>\(\sqrt{a}=\frac64=\frac32\)

=>\(a=\frac94\) (nhận)

uh là nhân

a) \(=\dfrac{254x\left(400-1\right)-145}{254+\left(400-1\right)x253}=\dfrac{254x400-254-145}{254+253x400-253}\)

\(=\dfrac{101600-399}{101200+1}=\dfrac{101211}{101201}=\dfrac{101201+10}{101201}=1+\dfrac{10}{101201}\)

b) \(=\dfrac{5392+\left(600+1\right)x5391}{5392x\left(600+1\right)-69}=\dfrac{5392+600x5391+5391}{5392x600+5392-69}\)

\(=\dfrac{10783+3234600}{3235200+5323}=\dfrac{\text{3245383}}{\text{3240523}}=\dfrac{3240523+60}{3240523}=1+\dfrac{60}{3240523}\)

c) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)

` ĐK:(-5)/(x^{2}+6)>=0`

Vì `-5<0` và `x^{2}+6>0`

`->(-5)/(x^{2}+6)<0`

Vậy căn thức trên không tồn tại, không có giá trị của `x` thỏa mãn

ĐKXĐ: \(x\in\varnothing\)

\(2,25:\dfrac{-6}{25}=\dfrac{9}{4}:\dfrac{-6}{25}\\ =-\dfrac{9}{4}\cdot\dfrac{25}{6}=-\dfrac{3}{4}\cdot\dfrac{25}{2}\\ =-\dfrac{75}{8}\)

\(2.25:\dfrac{-6}{25}\) = \(\dfrac{9}{4}:\dfrac{-6}{25}=\dfrac{9}{4}.\dfrac{25}{-6}=\dfrac{3}{4}.\dfrac{25}{-2}=\dfrac{75}{-8}=\dfrac{-75}{8}\)

a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)

\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)

\(\Rightarrow64x+96=72x-24\)

\(\Rightarrow8x=120\Rightarrow x=15\)

b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)

\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)

\(\Rightarrow221x-34=152x+380\)

\(\Rightarrow69x=414\Rightarrow x=6\)

\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)

\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)

\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)

\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)

\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)

mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3

a) 9/18 - (-7/12) + 13/32

= 13/12 + 13/32

= 143/96

b) (5/-8) + 14/25 - 6/10

= (-13/200) - 6/10

= -133/200

Chúc bạn học tốt!! ^^

a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)

\(=\dfrac{143}{96}\)

b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)

\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)

\(=\dfrac{-133}{200}\)