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

Ta có \(x-y+2z=-20\)

\(\Leftrightarrow x-y+\dfrac{1}{2}\cdot4z=-20\Leftrightarrow x-y+\dfrac{1}{2}\cdot3x=-20\)

\(\Leftrightarrow\dfrac{5}{2}x-y=-20\Leftrightarrow5x-2y=-40\)

Theo đề bài: \(\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{5x}{15}=\dfrac{2y}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{5x}{15}=\dfrac{2y}{10}=\dfrac{5x-2y}{15-10}=-\dfrac{40}{5}=-8\)

\(\Rightarrow\dfrac{x}{3}=-8\Rightarrow x=-24\)

\(\Rightarrow\dfrac{y}{5}=-8\Rightarrow y=-40\)

\(\Rightarrow3\cdot\left(-24\right)=4z\Rightarrow z=-18\)

\(\dfrac{x-1}{15}=\dfrac{60}{x-1}\) (ĐK: \(x\ne1\))

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=15\cdot60\)

\(\Leftrightarrow\left(x-1\right)^2=900\)

\(\Leftrightarrow\left(x-1\right)^2=30^2\)

\(\Leftrightarrow\left|x-1\right|=30\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)(T/m)

\(\dfrac{15}{2x-1}=\dfrac{5}{3}\) (ĐK: \(x\ne\dfrac{1}{2}\))

\(\Leftrightarrow3\cdot15=5\left(2x-1\right)\)

\(\Leftrightarrow45=10x-5\)

\(\Leftrightarrow10x=50\)

\(\Leftrightarrow x=5\) (T/m)

a) O là tâm hbh ABCD

\(\rightarrow\) O là trung điểm AC \(\rightarrow\overrightarrow{OA}+\overrightarrow{OC}=\overrightarrow{0}\)

    O là trung điểm BD \(\rightarrow\overrightarrow{OB}+\overrightarrow{OD}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OA}+\overrightarrow{OC}\right)+\left(\overrightarrow{OB}+\overrightarrow{OD}\right)=\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\) (đpcm)

b) Vì ABCD là hbh \(\rightarrow\overrightarrow{AB}=\overrightarrow{DC};\overrightarrow{AD}=\overrightarrow{BC}\)

Vì E là trung điểm AD \(\rightarrow\overrightarrow{EA}=\overrightarrow{DE}\)

 \(2\overrightarrow{CE}=\overrightarrow{CA}+\overrightarrow{CD}\rightarrow2\overrightarrow{EC}=-\left(\overrightarrow{CA}+\overrightarrow{CD}\right)=\overrightarrow{AC}+\overrightarrow{DC}\)

Ta có: \(\overrightarrow{EA}+\overrightarrow{EB}+2\overrightarrow{EC}=\overrightarrow{DE}+\overrightarrow{EB}+\overrightarrow{AC}+\overrightarrow{DC}\)

\(=\overrightarrow{DB}+\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AB}=2\overrightarrow{AB}+\overrightarrow{DC}\)

\(=2\overrightarrow{AB}+\overrightarrow{AB}=3\overrightarrow{AB}\) (đpcm)

c) Sửa đề : CM: vectơ AB + 2 lần vectơ EA + 4 lần vectơ ED = vectơ AC

Vì E là trung điểm AD \(\rightarrow\overrightarrow{ED}=\overrightarrow{AE};\overrightarrow{AD}=2\overrightarrow{AE}\)

\(\overrightarrow{AB}+2\overrightarrow{EA}+4\overrightarrow{ED}=\overrightarrow{AB}-2\overrightarrow{AE}+4\overrightarrow{AE}=\overrightarrow{AB}+2\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\)

Ta có: MB=3MC \(\rightarrow MB=\dfrac{3}{4}BC\)

mà M nằm giữa B và C \(\rightarrow\overrightarrow{BM}=\dfrac{3}{4}\overrightarrow{BC}\)

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\dfrac{3}{4}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\)