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

Bài 2: 

14: =(x-5)(x+1)

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?

Câu II: 

a) \(\sqrt{2x-1}=\sqrt{5}\left(x\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=5+1\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}\)

\(\Leftrightarrow x=3\left(tm\right)\)

b) \(6\sqrt{x-5}+\sqrt{9x-45}-2\sqrt{4x-20}=25\left(x\ge5\right)\)

\(\Leftrightarrow6\sqrt{x-5}+\sqrt{9\left(x-5\right)}-2\sqrt{4\left(x-5\right)}=25\)

\(\Leftrightarrow6\sqrt{x-5}+3\sqrt{x-5}-2\cdot2\sqrt{x-5}=25\)

\(\Leftrightarrow9\sqrt{x-5}-4\sqrt{x-5}=25\)

\(\Leftrightarrow5\sqrt{x-5}=25\)

\(\Leftrightarrow\sqrt{x-5}=5\)

\(\Leftrightarrow x-5=5^2\)

\(\Leftrightarrow x-5=25\)

\(\Leftrightarrow x=30\left(tm\right)\)

Bài III:

a: 

b: phương trình hoành độ giao điểm là:

2x-4=-x+2

=>2x+x=4+2

=>3x=6

=>x=6/3=2

Thay x=2 vào y=-x+2, ta được:

\(y=-2+2=0\)

Vậy: A(2;0)

c: Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=-x+2=-0+2=2\end{matrix}\right.\)

Vậy: B(0;2)

Tọa độ C là:

\(\left\{{}\begin{matrix}x=0\\y=2x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\cdot0-4=-4\end{matrix}\right.\)

Vậy: C(0;-4)

Ta có: A(2;0); B(0;2); C(0;-4)

\(AB=\sqrt{\left(0-2\right)^2+\left(2-0\right)^2}=2\sqrt{2}\)

\(AC=\sqrt{\left(0-2\right)^2+\left(-4-0\right)^2}=\sqrt{4^2+2^2}=2\sqrt{5}\)

\(BC=\sqrt{\left(0-0\right)^2+\left(-4-2\right)^2}=6\)

Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{8+20-36}{2\cdot2\sqrt{2}\cdot2\sqrt{5}}=\dfrac{-8}{8\sqrt{5}}=-\dfrac{1}{\sqrt{5}}\)

=>\(sinBAC=\sqrt{1-cos^2BAC}=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt{5}}\cdot2\sqrt{2}\cdot2\sqrt{5}=4\sqrt{2}\)

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:

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

Câu 1: A

Câu 2: D

Câu 3: B

Câu 4: C

Bài 1:

a) 

\(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{(\sqrt{x}-2)(\sqrt{x}+2)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right)\)

\(=\frac{4\sqrt{x}(\sqrt{x}-2)-8x}{(\sqrt{x}-2)(\sqrt{x}+2)}:\frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}=\frac{-4x-8\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{-\sqrt{x}+3}\)

\(=\frac{-4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}(\sqrt{x}-2)}{3-\sqrt{x}}=\frac{-4x(\sqrt{x}-2)}{(\sqrt{x}-2)(3-\sqrt{x})}=\frac{4x}{\sqrt{x}-3}\)

b)

Ta có:
\(m(\sqrt{x}-3).A>x+2025\)

\(\Leftrightarrow 4xm>x+2025\Leftrightarrow x(4m-1)>2025\)

\(\Leftrightarrow 4m-1>\frac{2025}{x}\Leftrightarrow m>\frac{1}{4}(\frac{2025}{x}+1)\) với mọi $x>9$

\(\Leftrightarrow m> \max \frac{1}{4}(\frac{2025}{x}+1), \forall x>9\Leftrightarrow m>56,5\)

Bài 2:

a) 

\(\left\{\begin{matrix} 4x-3y=19\\ 2x+3y=11\end{matrix}\right.\Rightarrow 6x=30\Rightarrow x=5\)

\(y=\frac{4x-19}{3}=\frac{4.5-19}{3}=\frac{1}{3}\)

b) 

\(3x^2-20x+12=0\Leftrightarrow (x-6)(3x-2)=0\)

\(\Rightarrow \left[\begin{matrix} x=6\\ x=\frac{2}{3}\end{matrix}\right.\)

\(m_{H_2O}=1,62\left(g\right)\Rightarrow n_{H_2O}=0,09\left(mol\right)\Rightarrow n_H=0,18\left(mol\right);m_H=0,18.1=0,18\left(g\right)\\ n_{CO_2}=\dfrac{2,64}{44}=0,06\left(mol\right)\Rightarrow n_C=n_{CO_2}=0,06\left(mol\right);m_C=0,06.12=0,72\left(g\right)\\ Vây:m_C+m_H=0,72+0,18=0,9< 1,38\\ \Rightarrow X.có.chứa.O\\ m_O=1,38-0,9=0,48\left(g\right);n_O=\dfrac{0,48}{16}=0,03\left(mol\right)\\ Đặt.X:C_aH_bO_c\left(a,b,c:nguyên,dương\right)\\ Ta.có:a:b:c=0,06:0,18:0,03=2:6:1\\ \Rightarrow CTĐG:C_2H_6O\\ M_X=23.2=46\left(\dfrac{g}{mol}\right)=M_{C_2H_6O}\\ \Rightarrow X:C_2H_6O\)