Bài 8.9.10 của câu 14 hay 15 bạn?

Câu 15:

1: Ta có: \(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{1+\sqrt{2}}\)

\(=\dfrac{1+\sqrt{2}-1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)

\(=-2\sqrt{2}\)

2: Ta có: \(\dfrac{1}{\sqrt{5}+1}+\dfrac{1}{\sqrt{5}-1}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)

\(=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\)

Câu 14:

8: Ta có: \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}-2}\)

9: Ta có: \(\dfrac{3\sqrt{x}+2}{2\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+4}-\dfrac{x-6\sqrt{x}+5}{2x-7\sqrt{x}-4}\)

\(=\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}+4\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\dfrac{x-6\sqrt{x}+5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\dfrac{3x+12\sqrt{x}+2\sqrt{x}+8+2x-\sqrt{x}-2\sqrt{x}+1-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\dfrac{4x+20\sqrt{x}+4}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

10 nhân 10 =100 nha em !

100 nha

one hundred nhé em

có nghĩa là 100 nè

Câu 1: \(\frac{\pi}{2}<\alpha,\beta<\pi\)

=>\(\sin\alpha>0;\sin\beta>0;cos\alpha<0;cos\beta<0\)

\(\sin^2\alpha+cos^2\alpha=1\)

=>\(cos^2\alpha=1-\sin^2\alpha=1-\left(\frac13\right)^2=\frac89\)

mà \(cos\alpha<0\)

nên \(cos\alpha=-\frac{2\sqrt2}{3}\)

Ta có: \(\sin^2\beta+cos^2\beta=1\)

=>\(\sin^2\beta=1-\left(-\frac23\right)^2=1-\frac49=\frac59\)

mà \(\sin\beta>0\)

nên \(\sin\beta=\frac{\sqrt5}{3}\)

\(\sin\left(\alpha+\beta\right)=\sin\alpha\cdot cos\beta+cos\alpha\cdot\sin\beta\)

\(=\frac13\cdot\frac{-2}{3}+\frac{-2\sqrt2}{3}\cdot\frac{\sqrt5}{3}=\frac{-\sqrt2-2\sqrt{10}}{9}\)

Câu 2:

\(P=cos\left(a+b\right)\cdot cos\left(a-b\right)\)

\(=\frac12\cdot\left\lbrack cos\left(a+b+a-b\right)+cos\left(a+b-a+b\right)\right\rbrack=\frac12\cdot\left\lbrack cos2a+cos2b\right\rbrack\)

\(=\frac12\cdot\left\lbrack2\cdot cos^2a-1+2\cdot cos^2b-1\right\rbrack=cos^2a+cos^2b-1\)

\(=\left(\frac13\right)^2+\left(\frac14\right)^2-1=\frac19+\frac{1}{16}-1=\frac{25}{144}-1=-\frac{119}{144}\)

A) 7,2 - 0,25 + x = 10

6,95 + x = 10

x = 10 - 6,95

x = 3,05

B) 20 - 4,5 - x = 12,75

15,5 - x = 12,75

x = 15,5 - 12,75

x = 2,75

Đề bài yêu cầu gì?

=> (x+x+...+x) + (1+2+...+10)=241

   .....10x.....      

=> 10x + [ (10+1).10 :2] = 240

=> 10x + 55 = 240

=> 10x        = 240 - 55 = 185

=> x            = 185 : 10 = 18,5

(x+1) + (x+2)+(x+3)+ ....+(x+9)+(x+10) = 240

(x+x+x+...+x) + (1+2+3+4+...+9+10) = 240

        10x        +             55              = 240

                     10x                            = 240 - 55

                     10x                            = 185

                       x                             = 185 : 10

                       x                             = 18,5

Vậy ...

Cách tính :

Số số hạng là :

(10-1):1=1=10 (số)

Tổng số hạng là :

[10x[10+1]:1+1] :2 =60

Vậy x là :

(240-60): 10=18

        Đáp số:18

là sao

Môn tin học lớp 10 ý bạn..

Bạn lớp mấy?.😪😪

bnj thi tin à