6.23

a)\(\dfrac{-5}{3}\) - \(\dfrac{-7}{3}\)

=\(\dfrac{-5-\left(-7\right)}{3}\)

=\(\dfrac{2}{3}\)

b)\(\dfrac{5}{6}\)- \(\dfrac{8}{9}\) 

=\(\dfrac{5.3}{6.3}\)- \(\dfrac{8.2}{9.2}\)

=\(\dfrac{15}{18}\)-\(\dfrac{16}{18}\)

=\(\dfrac{-1}{18}\)

6.24 

A=(-\(\dfrac{3}{11}\)) + \(\dfrac{11}{8}\) - \(\dfrac{3}{8}\) + (-\(\dfrac{8}{11}\))

= [ (-\(\dfrac{3}{11}\)) +  (-\(\dfrac{8}{11}\)) ] + ( \(\dfrac{11}{8}\) - \(\dfrac{3}{8}\) )

=\(\dfrac{-11}{11}\) + \(\dfrac{8}{8}\)

= -1 + 1

=0

Tổng phần lương đã chi tiêu và mua quà là:
\(\dfrac{2}{5}\) + \(\dfrac{1}{4}\) = \(\dfrac{8}{20}\) + \(\dfrac{5}{20}\)=\(\dfrac{13}{20}\)

Phần tiền lương còn lại :

1 - \(\dfrac{13}{20}\) = \(\dfrac{20}{20}\) - \(\dfrac{13}{20}\) = \(\dfrac{7}{20}\)

C1 = 7,5 : 0,75 = 10

C2 = 6.24 :0,75 + 1.26 : 0.75 =10

Bài 9:

Ta có: \(a-4+\sqrt{16-8a+a^2}\)

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

\(=a-4+a-4\)

=2a-8

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

\(\dfrac{x^2-2x\sqrt{15}+15}{x^2-15}=\dfrac{\left(x-\sqrt{15}\right)^2}{\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)}=\dfrac{x-\sqrt{15}}{x+\sqrt{15}}\)

\(\dfrac{4x^2-6}{4x^2-4x\sqrt{6}+6}=\dfrac{\left(2x-\sqrt{6}\right)\left(2x+\sqrt{6}\right)}{\left(2x-\sqrt{6}\right)^2}=\dfrac{2x+\sqrt{6}}{2x-\sqrt{6}}\)

b) \(\dfrac{a^2+2a\sqrt{8}+8}{a^2-8}=\dfrac{\left(a+\sqrt{8}\right)^2}{\left(a+\sqrt{8}\right)\left(a-\sqrt{8}\right)}=\dfrac{a+\sqrt{8}}{a-\sqrt{8}}\)

\(\dfrac{9x^2-15}{9x^2-6x\sqrt{15}+15}=\dfrac{\left(3x-\sqrt{15}\right)\left(3x+\sqrt{15}\right)}{\left(3x-\sqrt{15}\right)^2}=\dfrac{3x+\sqrt{15}}{3x-\sqrt{15}}\)

\(\dfrac{a^2-2a\sqrt{7}+7}{a^2-7}=\dfrac{\left(a-\sqrt{7}\right)^2}{\left(a-\sqrt{7}\right)\left(a+\sqrt{7}\right)}=\dfrac{a-\sqrt{7}}{a+\sqrt{7}}\)

ai giúp mình đi mọi người

1:

Khi x=36 thì \(A=\dfrac{6+4}{6-4}=\dfrac{10}{2}=5\)

2: \(B=\dfrac{3\sqrt{x}-12+16-2\sqrt{x}}{x-16}=\dfrac{\sqrt{x}+4}{x-16}=\dfrac{1}{\sqrt{x}-4}\)

3: A=B*|x-2|

=>|x-2|=A/B=căn x+4

=>x-2=căn x+4 hoặc x-2=-căn x-4

=>x-căn x-6=0 hoặc x+căn x+2=0

=>x=9

=>x+1/2+x+1/6+...+x+1/90=99,9

=>\(\left(x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=99.9\)

=>\(9x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=99.9\)

=>9x+(1-0,1)=99,9

=>9x=99,9+0,1-1=100-1=99

=>x=11

Bài 4:

a. \(\left\{{}\begin{matrix}R1=\dfrac{U1^2}{P1}=\dfrac{120^2}{40}=360\Omega\\R2=\dfrac{U2^2}{P2}=\dfrac{120^2}{60}=240\Omega\end{matrix}\right.\)

b. \(P1< P2=>\) đèn 2 sáng hơn.

c. \(U1+U2=120+120=240V=U=240V=>\) đèn sáng bình thường.

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