Bài 1:

a: \(\dfrac{a}{b}>1\)

=>\(\dfrac{a}{b}-1>0\)

=>\(\dfrac{a-b}{b}>0\)

mà b>0

nên a-b>0

=>a>b

b: a>b

=>\(\dfrac{a}{b}>\dfrac{b}{b}\)

=>\(\dfrac{a}{b}>1\)

c: a/b<1

=>\(\dfrac{a}{b}-1< 0\)

=>\(\dfrac{a-b}{b}< 0\)

mà b>0

nên a-b<0

=>a<b

d: a<b

=>\(\dfrac{a}{b}< \dfrac{b}{b}\)

=>\(\dfrac{a}{b}< 1\)

1. Where do they live?

2. What do those girls sell (there)?

3. What time do they get home (every night)?

4. What language does she speak well?

1 Where do they live?

2 What do those girls sell there?

3 What time do they get home every night?

4 What language does she speak well?

\(2x^2-6x+1=0\)

\(\Leftrightarrow4x^2-12x+2=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x.3+9=7\)

\(\Leftrightarrow\left(2x-3\right)^2=7\)

\(\Leftrightarrow2x-3=\pm\sqrt{7}\)

\(\Leftrightarrow2x=\pm\sqrt{7}+3\)

\(\Leftrightarrow x=\dfrac{\pm\sqrt{7}+3}{2}\)

Vậy ...

`2x^2 - 6x + 1 = 0`

`Δ' = \(\left(\dfrac{b}{2}\right)^2-ac\) = 3^2 - 2.1 = 7 > 0`

=> Phương trình có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{-\dfrac{b}{2}+\sqrt{\Delta}}{2}=\dfrac{3+\sqrt{7}}{2}\\x=\dfrac{-\dfrac{b}{2}-\sqrt{\Delta}}{2}=\dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\)

Vậy ....

\(\left(x^2-9\right)-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x-3\right)\left[\left(x+3\right)-9\left(x-3\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-9x+27\right)=0\\ \Leftrightarrow\left(x-3\right)\left(30-8x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\30-8x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{15}{4}\end{matrix}\right.\)

`#3107.101107`

\(\left(x^2-9\right)-9\left(x-3\right)^2=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-9\left(x-3\right)^2=0\\ \Rightarrow\left(x-3\right)\left[x+3-9\left(x-3\right)\right]=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3-9x+27=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\-8x+30=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\-8x=-30\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\8x=30\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{15}{4}\end{matrix}\right.\)

Vậy, \(x\in\left\{3;\dfrac{15}{4}\right\}.\)

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Các HĐT sử dụng trong bài:

\(\left(A-B\right)^2=A^2-2AB+B^2\\ A^2-B^2=\left(A-B\right)\left(A+B\right).\)

\(\left(x^2-9\right)-9\left(x-3\right)^2\\ =\left(x-3\right)\left(x+3\right)-9\left(x-3\right)^2\\ =\left(x-3\right)\left[\left(x+3\right)-9\left(x-3\right)\right]\\ =\left(x-3\right)\left(x+3-9x+27\right)\\ =\left(x-3\right)\left(30-8x\right)\)

I exercise regularly so as to improve my mental health.

She goes to the clinic when needed

We avoid the city because of the exhaust fumes.

The farmer plants seeds after preparing the agricultural fields.

She completed her vocational training before starting her job.

Despite losing his job, he kept looking for work because of unemployment.

He studies hard when learning about economic principles.

We distribute food to the needy after collecting donations.

She donates money to the poor to support charity.

She learns how to recycle plastic products because of worrying about global warming.

They moved to a new area to escape overpopulation.

She attends workshops after finishing her vocational training.

Số bị chia là \(17\cdot50+2=852\)

=>Chọn A

a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC=\sqrt{5^2+12^2}=13\left(cm\right)\)

Xét ΔABC có AD là phân giác

nên \(\dfrac{BD}{AB}=\dfrac{CD}{AC}\)

=>\(\dfrac{BD}{5}=\dfrac{CD}{12}\)

mà BD+CD=BC=13cm

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{BD}{5}=\dfrac{CD}{12}=\dfrac{BD+CD}{5+12}=\dfrac{13}{17}\)

=>\(BD=\dfrac{13}{17}\cdot5=\dfrac{65}{17}\left(cm\right);CD=\dfrac{13}{17}\cdot12=\dfrac{156}{17}\left(cm\right)\)

b: Xét ΔCDE vuông tại D và ΔCAB vuông tại A có

\(\widehat{DCE}\) chung

Do đó: ΔCDE~ΔCAB

=>\(k=\dfrac{CD}{CA}=\dfrac{156}{17}:12=\dfrac{13}{17}\)

c: ΔCDE~ΔCAB

=>\(\dfrac{CD}{CA}=\dfrac{CE}{CB}\)

=>\(\dfrac{CD}{CE}=\dfrac{CA}{CB}\)

Xét ΔCDA và ΔCEB có

\(\dfrac{CD}{CE}=\dfrac{CA}{CB}\)

\(\widehat{C}\) chung

Do đó: ΔCDA~ΔCEB

=>\(\dfrac{DA}{EB}=\dfrac{CA}{CB}\)

=>\(DA\cdot CB=BE\cdot AC\)

d: ΔCDE~ΔCAB

=>\(\dfrac{DE}{AB}=\dfrac{CD}{CA}\)

=>\(\dfrac{DE}{5}=\dfrac{156}{17}:12=\dfrac{13}{17}\)

=>\(DE=\dfrac{13}{17}\cdot5=\dfrac{65}{17}\left(cm\right)\)

Xét tứ giác ABDE có \(\widehat{EAB}+\widehat{EDB}=90^0+90^0=180^0\)

nên ABDE là tứ giác nội tiếp

=>\(\widehat{DEB}=\widehat{DAB}=45^0\)

Xét ΔDEB vuông tại D có \(\widehat{DEB}=45^0\)

nên ΔDEB vuông cân tại D

ΔBDE vuông cân tại D

=>\(S_{BDE}=\dfrac{1}{2}\cdot DB\cdot DE=\dfrac{1}{2}\cdot DB^2=\dfrac{1}{2}\cdot\left(\dfrac{65}{17}\right)^2=\dfrac{1}{2}\cdot\dfrac{4225}{289}=\dfrac{4225}{578}\left(cm^2\right)\)