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a: Xét ΔADH vuông tại H và ΔBDA vuông tại A có
\(\widehat{ADH}\) chung
Do đó: ΔADH∼ΔBDA
b: Xét ΔHAD vuông tại H và ΔHBA vuông tại H có
\(\widehat{HAD}=\widehat{HBA}\)
Do đó: ΔHAD∼ΔHBA
Suy ra: HA/HB=HD/HA
hay \(HA^2=HB\cdot HD\)
a) Xét \(\Delta ADH\) và \(\Delta BDA:\)
\(\widehat{H}=\widehat{A}\left(=90^o\right).\)
\(\widehat{D}\) chung.
\(\Rightarrow\Delta ADH\sim\Delta BDA\left(g-g\right).\)
b) Xét \(\Delta BDA\) và \(\Delta BAH:\)
\(\widehat{BAD}=\widehat{BHA}\left(=90^o\right).\)
\(\widehat{B}\) chung.
\(\Rightarrow\Delta BDA\sim\) \(\Delta BAH\left(g-g\right).\)
Mà \(\Delta ADH\sim\Delta BDA\left(cmt\right).\)
\(\Rightarrow\Delta ADH\sim\Delta BAH.\)
\(\Rightarrow\dfrac{AH}{BH}=\dfrac{DH}{AH}\) (2 cạnh tương ứng).
\(\Rightarrow AH^2=DH.BH.\)
Bài 2:
\(a,=x\left(x+7\right)\\ b,=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
Bài 3:
\(a,\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Bài 4:
\(a,\widehat{D}=360^0-\widehat{A}-\widehat{B}-\widehat{C}=150^0\)
\(b,\dfrac{x+10}{2}=12\Leftrightarrow x+10=24\Leftrightarrow x=14\left(cm\right)\)
2:
Gọi số sách lúc đầu ở tủ 1 và tủ 2 lần lượt là a,b
Theo đề, ta có:
a+b=600 và a-80=1/2(b+80)
=>a=280 và b=320
a.\(A=\left(\dfrac{x^2-3}{x^2-9}+\dfrac{1}{x-3}\right):\dfrac{x}{x+3}\)
\(A=\left(\dfrac{x^2-3+\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x}{x+3}\)
\(A=\left(\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+3}{x}\)
\(A=\dfrac{\left(x^2+x\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(x+1\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x+1}{x-3}\)
b.\(A=3\)
\(\Leftrightarrow\dfrac{x+1}{x-3}=3\)
\(\Leftrightarrow\dfrac{x+1}{x-3}=\dfrac{3\left(x-3\right)}{x-3}\)
\(\Leftrightarrow x+1=3\left(x-3\right)\)
\(\Leftrightarrow x+1=3x-9\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\left(tm\right)\)
Vậy \(x=5\) thì \(A=3\)
`[x+1]/2021+[x+2]/2020+[x+3]/2019+[x+4]/2018=-4`
`<=>[x+1]/2021+1+[x+2]/2020+1+[x+3]/2019+1+[x+4]/2018+1=-4+4`
`<=>[x+1+2021]/2021+[x+2+2020]/2020+[x+3+2019]/2019+[x+4+2018]/2018=0`
`<=>[x+2022]/2021+[x+2022]/2020+[x+2022]/2019+[x+2022]/2018=0`
`<=>(x+2022)(1/2021+2020+1/2019+1/2018)=0`
Mà `1/2021+2020+1/2019+1/2018 \ne 0`
`=>x+2022=0`
`<=>x=-2022`
Vậy `S={-2022}`
4: \(D=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(A=\left(x^2-6x+9\right)-7=\left(x-3\right)^2-7\ge7\\ A_{min}=7\Leftrightarrow x=3\\ B=\left(9x^2+6x+1\right)-4=\left(3x+1\right)^2-4\ge-4\\ B_{min}=-4\Leftrightarrow x=-\dfrac{1}{3}\\ C=\left(x^2-2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{9}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\\ C_{min}=-\dfrac{9}{4}\Leftrightarrow x=\dfrac{5}{2}\\ D=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ D_{min}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(E=3\left(x^2+2\cdot\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\\ E_{min}=-\dfrac{4}{3}\Leftrightarrow x=-\dfrac{1}{3}\\ F=x^2-2x+1+x^2-4x+4+2021\\ F=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{4031}{2}=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{4031}{2}\ge\dfrac{4031}{2}\\ F_{min}=\dfrac{4031}{2}\Leftrightarrow x=\dfrac{3}{2}\)
Bài 2:
1: ĐKXĐ: x<>1
\(\dfrac{x}{x-1}+\dfrac{1}{1-x}\)
\(=\dfrac{x}{x-1}-\dfrac{1}{x-1}\)
\(=\dfrac{x-1}{x-1}=1\)
2: ĐKXĐ: x<>3/2
\(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)
\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)
\(=\dfrac{11x+x-18}{2x-3}=\dfrac{12x-18}{2x-3}\)
\(=\dfrac{6\left(2x-3\right)}{2x-3}\)
=6
3: ĐKXĐ: x<>1/2
\(\dfrac{4x+5}{2x-1}+\dfrac{5-9x}{1-2x}\)
\(=\dfrac{4x+5}{2x-1}+\dfrac{9x-5}{2x-1}\)
\(=\dfrac{4x+5+9x-5}{2x-1}=\dfrac{13x}{2x-1}\)
4: ĐKXĐ: x<>2/5
\(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)
\(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)
5: ĐKXĐ: \(x\ne\pm y\)
\(\dfrac{xy}{x^2-y^2}-\dfrac{x^2}{y^2-x^2}\)
\(=\dfrac{xy}{x^2-y^2}+\dfrac{x^2}{x^2-y^2}\)
\(=\dfrac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x-y}\)
6: ĐKXĐ: \(x\notin\left\{0;7\right\}\)
\(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(7-x\right)}\)
\(=\dfrac{4x+13}{5x\left(x-7\right)}+\dfrac{x-48}{5x\left(x-7\right)}\)
\(=\dfrac{4x+13+x-48}{5x\left(x-7\right)}\)
\(=\dfrac{5x-35}{x\left(5x-35\right)}=\dfrac{1}{x}\)
7: ĐKXĐ: \(x\ne1\)
\(\dfrac{x+2}{x-1}-\dfrac{x-9}{1-x}-\dfrac{x-9}{1-x}\)
\(=\dfrac{x+2}{x-1}+\dfrac{x-9}{x-1}+\dfrac{x-9}{x-1}\)
\(=\dfrac{x+2+x-9+x-9}{x-1}=\dfrac{3x-16}{x-1}\)
8: ĐKXĐ:x<>1
\(\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
\(=\dfrac{2x^2-x}{x-1}-\dfrac{x+1}{x-1}+\dfrac{2-x^2}{x-1}\)
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}\)
=x-1
9: ĐKXĐ: x<>3
\(\dfrac{4-x^2}{x-3}+\dfrac{2x-x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}+\dfrac{x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2+x^2-2x+5-4x}{x-3}=\dfrac{-6x+9}{x-3}\)
10: ĐKXĐ: x<>5
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{5-x}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1}{x-5}-\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1-x+18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)
1: \(\dfrac{15x}{7y^3}\cdot\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x\cdot2y^2}{7y^3\cdot x^2}=\dfrac{30xy^2}{7x^2y^3}=\dfrac{30}{7xy}\)
2: \(\dfrac{6x^3}{7y^4}\cdot\dfrac{35y^2}{24x}\)
\(=\dfrac{6x^3}{24x}\cdot\dfrac{35y^2}{7y^4}\)
\(=\dfrac{x^2}{4}\cdot\dfrac{5}{y^2}=\dfrac{5x^2}{4y^2}\)
3: \(\dfrac{4y^2}{x^4}\cdot\dfrac{-3x^2}{8y}\)
\(=\dfrac{4y^2}{8y}\cdot\dfrac{-3x^2}{x^4}=\dfrac{y}{2}\cdot\dfrac{-3}{x^2}=\dfrac{-3y}{2x^2}\)
4: \(\dfrac{-18y^3}{25x^4}:\dfrac{9y^3}{-15x^2}\)
\(=\dfrac{18y^3}{25x^4}\cdot\dfrac{15x^2}{9y^3}\)
\(=\dfrac{18y^3}{9y^3}\cdot\dfrac{15x^2}{25x^4}=2\cdot\dfrac{3}{5x^2}=\dfrac{6}{5x^2}\)
5: \(\dfrac{8y^2}{9x^2}:\dfrac{4y}{3x^2}\)
\(=\dfrac{8y^2}{9x^2}\cdot\dfrac{3x^2}{4y}=\dfrac{8y^2}{4y}\cdot\dfrac{3x^2}{9x^2}=\dfrac{1}{3}\cdot2y=\dfrac{2y}{3}\)
6: \(\dfrac{-20x}{3y^2}:\dfrac{-4x^3}{5y}\)
\(=\dfrac{20x}{3y^2}:\dfrac{4x^3}{5y}\)
\(=\dfrac{20x}{3y^2}\cdot\dfrac{5y}{4x^3}=\dfrac{20x}{4x^3}\cdot\dfrac{5y}{3y^2}=\dfrac{5}{3y}\cdot\dfrac{5}{x^2}=\dfrac{25}{3x^2y}\)
7: \(\dfrac{\left(x+4\right)^2}{4x+12}:\dfrac{x+4}{3x+9}\)
\(=\dfrac{\left(x+4\right)^2}{4\left(x+3\right)}:\dfrac{x+4}{3\left(x+3\right)}\)
\(=\dfrac{\left(x+4\right)^2}{4\left(x+3\right)}\cdot\dfrac{3\left(x+3\right)}{x+4}=\dfrac{3\left(x+4\right)}{4}\)
8: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}\)
\(=\dfrac{5\cdot\left(-2\right)}{4}=-\dfrac{10}{4}=-\dfrac{5}{2}\)
9: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
10: \(\dfrac{x^2-4}{x^2-x}:\dfrac{x^2+2x}{x-1}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-1\right)}\cdot\dfrac{x-1}{x\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)}{x^2\left(x+2\right)}\)
