Đề dài thế này sao giải thích nhanh cho e đc

Part 1

1 C

2 B

3 D

4 C

5 B

6 A

Part 2

1 T

2 F

3 F

4 F

V

1 That old house has just been bought

2 If he doesn't take these pills, he won't be better

3 I suggest taking a train

4 Spending the weekend in the countryside is very wonderful

nhiều thật đấy

no khó lắm tui lớp 5 thôi chịu

:(((((((((((

1.a

\(\lim\limits_{x\rightarrow2}\dfrac{x^3+3x^2-9x-2}{x^3-x-6}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x^2+5x+1\right)}{\left(x-2\right)\left(x^2+2x+3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+5x+1}{x^2+2x+3}=\dfrac{15}{11}\)

b.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x+3}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-x+3}{\sqrt{x^2-x+3}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1+\dfrac{3}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{3}{x^2}}-1}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

2.

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{2x-\sqrt{x^2+3}}{x-1}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{4x^2-\left(x^2+3\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x^2-1\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x+1\right)}{2x+\sqrt{x^2+3}}\)

\(=\dfrac{3.2}{2+\sqrt{4}}=\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{2x+7}{6}=\dfrac{2.1+7}{6}=\dfrac{3}{2}\)

\(f\left(1\right)=\dfrac{2.1+7}{6}=\dfrac{3}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)

\(\Rightarrow\) Hàm liên tục tại \(x=1\)

Đề bài đâu em hi?

Đề 1:

Bài 1:

\(a,=\sqrt{\left(\sqrt{7}+1\right)^2}-\left|-1+\sqrt{7}\right|=\sqrt{7}+1-\sqrt{7}+1=2\\ b,=2\sqrt{2}-4\sqrt{2}-5\sqrt{2}+\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}-7\sqrt{2}=\dfrac{-13\sqrt{2}}{\sqrt{2}}\)

Bài 2:

\(PT\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=\dfrac{1}{2}\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2}=1\\x=-\dfrac{1}{2}+\dfrac{1}{2}=0\end{matrix}\right.\)

Bài 3:

\(a,M=\dfrac{a-2\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{2\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}=\dfrac{2}{\sqrt{a}+1}\\ b,M< 1\Leftrightarrow\dfrac{2}{\sqrt{a}+1}-1< 0\Leftrightarrow\dfrac{1-\sqrt{a}}{\sqrt{a}+1}< 0\\ \Leftrightarrow1-\sqrt{a}< 0\left(\sqrt{a}+1>0\right)\\ \Leftrightarrow a>1\)

2:

a: pi/2<a<pi

=>cosa<0

sin^2a+cos^2a=1

=>cos^2a=1-4/9=5/9

=>cosa=-căn 5/3

cos2a=2*cos^2a-1=2*5/9-1=10/9-1=1/9

sin(2a-pi/6)

=sin2a*cospi/6-cos2a*sinpi/6

=2*sina*cosa*(căn 3/2)-1/9*1/2

\(=2\cdot\dfrac{2}{3}\cdot\dfrac{-\sqrt{5}}{3}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{18}=\dfrac{-4\sqrt{15}-1}{18}\)

b; tan a=2

=>sin a=2*cosa

\(A=\dfrac{3\cdot\left(2\cdot cosa\right)^2-cos^2a+2}{5\cdot\left(2\cdot cosa\right)^2+3cosa\cdot2cosa}\)

\(=\dfrac{12\cdot cos^2a-cos^2a+2}{20cos^2a+6cos^2a}\)

\(=\dfrac{11cos^2a+2\left(4cos^2a+cos^2a\right)}{26cos^2a}=\dfrac{21}{26}\)

4:

a: (C): x^2+y^2-4x+2y-4=0

=>x^2-4x+4+y^2+2y+1=9

=>(x-2)^2+(y+1)^2=9

=>I(2;-1); R=3

b: Gọi (d) là phương trình cần tìm

(d)//4x+3y-1=0

=>(d): 4x+3y+c=0

I(2;-1);R=3

Theo đề, ta có: d(I;(d))=R=3

=>\(\dfrac{\left|4\cdot2+3\cdot\left(-1\right)+c\right|}{\sqrt{4^2+3^2}}=3\)

=>|c+5|=15

=>c=10 hoặc c=-20

1A 2D 3A 4C 5A 6D 7C 8A 9D 10C 11A 12A 13D 14C 15D 16C 17A 18B 19B 20A 21B 22C 23C 24D 25D 26C 27D 28C 29B

123 + [ 54 + ( - 123 ) + 46 ]

= 123 + [ - 69 + 46 ]

= 123 + ( - 23 )

= 100 

TK

Bài làm

123 + [ 54 + ( -123 ) + 46 ]

= 123 + [ 54 - 123 + 46 ]

= 123 - 23

= 100

123 + [54 + (-123) + 46]

=123+54+(-123)+46

=[123+(-123)]+(54+46)

=0+100

=100

Bạn cần bài nào ạ? Nếu bạn cần giúp tất cả thì bạn tách ra từng CH khác nhau nhé!

Câu 10: B

B