1.A  2.B  3.C  4.A  5.D  6.C  7.D  8.B  9.C  10.A  11.B

\(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{2\sqrt{x}}{1-x}\) (ĐK: \(x\ge0;x\ne1\))

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

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

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

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

a) Ta có: \(\widehat{ABH}+\widehat{HBz}=180^o\) (2 góc kề bù)

\(\rightarrow\widehat{ABH}+150^o=180^o\rightarrow\widehat{ABH}=30^o\)

mà \(\widehat{xAB}=30^o\) ;2 góc \(\widehat{ABH}\) và \(\widehat{xAB}\) ở vị trí so le trong với nhau

\(\Rightarrow\) xy//mn

b) Ta có : xy//mn

mà \(AH\perp mn\) \(\rightarrow AH\perp xy\) (T/c từ vuông góc tới song song)

\(\Rightarrow\widehat{HAy}=90^o\)

c) Ta có: \(\widehat{xAB}+\widehat{BAH}=\widehat{xAH}=90^o\left(AH\perp xy\right)\)

\(\rightarrow30^o+\widehat{BAH}=90^o\rightarrow\widehat{BAH}=60^o\left(1\right)\)

lại có \(\widehat{BAH}+\widehat{HAt}=180^o\) (2 góc kề bù)

\(\rightarrow60^o+\widehat{HAt}=180^o\rightarrow\widehat{HAt}=120^o\)

\(\rightarrow\widehat{HAv}=\dfrac{1}{2}\widehat{HAt}=\dfrac{1}{2}\cdot120^o=60^o\) (Av là phân giác góc HAt) (2)

Từ (1)(2) \(\rightarrow\widehat{BAH}=\widehat{HAv}=60^o\)

\(\Rightarrow\) AH là phân giác \(\widehat{BAv}\)

a) \(x^2-xz-9y^2+3yz=\left(x^2-9y^2\right)-z\left(x-3y\right)\)

\(=\left(x+3y\right)\left(x-3y\right)-z\left(x-3y\right)=\left(x-3y\right)\left(x+3y-z\right)\)

b) \(12x^3+4x^2-27xy-9y=4x^2\left(3x+1\right)-9y\left(3x+1\right)=\left(3x+1\right)\left(4x^2-9y\right)\)

c) \(x^4-25x^2+20x-4=x^4-\left(25x^2-20x+4\right)=x^4-\left(5x-2\right)^2=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)d) \(x^2y^2\cdot\left(x^2-9\right)-x^2+9=x^2y^2\cdot\left(x^2-9\right)-\left(x^2-9\right)=\left(x^2-9\right)\left(x^2y^2-1\right)\)

\(=\left(x-3\right)\left(x+3\right)\left(xy-1\right)\left(xy+1\right)\)

1) \(4-x^2-2xy-y^2=4-\left(x^2+2xy+y^2\right)=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)

2) \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)=25-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)3) \(4x^2+4xy+y^2-9=\left(4x^2+4xy+y^2\right)-9=\left(2x+y\right)^2-9=\left(2x+y-3\right)\left(2x+y+3\right)\)4)\(x^2+4xy-16+4y^2=\left(x^2+4xy+4y^2\right)-16=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)

\(x^2-\left(m+2\right)x+m=0\) (1)

a) Ta có: \(\Delta=\left[-\left(m+2\right)\right]^2-4\cdot1\cdot m=m^2+4m+4-4m=m^2+4>0\forall m\in R\)

\(\Rightarrow\) PT(1) luôn có 2 nghiệm pb x₁, x₂ với mọi m

b) Theo ĐL Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+2\\x_1x_2=\dfrac{c}{a}=m\end{matrix}\right.\)

Ta có: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{x_1+x_2-2}\)

\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{x_1+x_2-2}\)

\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m+2-2}\)

\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m}\) (ĐK: \(m\ne0\) )

\(\Leftrightarrow m\left(m+2\right)=m\)

\(\Leftrightarrow m^2+2m=m\)

\(\Leftrightarrow m^2+m=0\)

\(\Leftrightarrow m\left(m+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\left(L\right)\\m=-1\end{matrix}\right.\)

Vậy m = -1 là GT cần tìm

a) \(2x^2-5x+1=0\)(1)

Ta có: \(\Delta=\left(-5\right)^2-4\cdot2\cdot1=25-8=17>0\)

\(\Rightarrow\) PT (1) có 2 nghiệm phân biệt (đpcm)

b)Gọi x₁,x₂ là 2 nghiệm của PT(1)

 Theo ĐL Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{2}\\x_1x_2=\dfrac{c}{a}=\dfrac{1}{2}\end{matrix}\right.\)

\(A=x_1\left(x_1+2024\right)+x_2\left(x_2+2025\right)-x_2\)

\(=x_1^2+2024x_1+x_2^2+2025x_2-x_2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2+2024\left(x_1+x_2\right)\)

\(=\left(\dfrac{5}{2}\right)^2-2\cdot\dfrac{1}{2}+2024\cdot\dfrac{5}{2}\)

\(\dfrac{25}{4}-1+5060\)

\(=\dfrac{20261}{4}\)

\(a\left(b^2+c^2+bc\right)+b\left(c^2+a^2+ac\right)+c\left(a^2+b^2+ab\right)\)

\(=ab^2+ac^2+abc+bc^2+a^2b+abc+a^2c+b^2c+abc\)

\(=ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)+abc+abc+abc\)

\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\left(\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\right)\)

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)

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

\(=\sqrt{4-3}=1\)

1. How many

2. How many

3. How much

4. How many

5. How many

(Bài này điền từ hả bạn?)

B. 2-3-4-1