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1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
XIV.
1. Her sister is more beautiful than her
2. English exercises is more difficult than Literature exercises
3. My new class is bigger tham my old class
4. Phi is more handsome than anybody in the class
5. Loan is taller than anyone in the family
IV.
3. French is not as difficult as English
4. That house is bigger than this house
5. A new house is more expensive than an old house
6. Yesterday is not as hot as today
7. My sister is taller than me
8. Girls are not as strong as boys
X.
1. Lesson 1 is not as long as Lesson 2
2. Ha Noi is smaller than HCM City
3. Old English is not as difficult as new French
4. Quang is not as thin as Quan
5. Your computer is more expensive than my computer
XI.
3. Minh is thinner than Hoang
4. Driving a motorbike is not as difficult as driving a car
XII.
1. Boys are stronger than girls
2. I am more intelligent than my sister
3. A new house is more expensive than an old house
4. English is more difficult than French
5. That river is longer than this river
6. New films are more interesting than old films
XIII.
3. Quan is the most intelligent student in my school
4. HCM City is the biggest city in VN
a, Theo định lí Pytago tam giác HBM vuông tại B
\(HM=\sqrt{BH^2+BM^2}=17cm\)
Ta có \(S_{HBM}=\dfrac{1}{2}.BI.HM;S_{HBM}=\dfrac{1}{2}.BH.BM\)
\(\Rightarrow BI=\dfrac{BH.BM}{HM}=\dfrac{120}{17}cm\)
b, Xét tam giác HIB và tam giác HBM có
^H _ chung ; ^HIB = ^HBM = 900
Vậy tam giác HIB ~ tam giác HBM (g.g)
\(\dfrac{HI}{HB}=\dfrac{HB}{HM}\Rightarrow HI=\dfrac{HB^2}{HM}=\dfrac{225}{17}cm\)
c, Xét tam giác MIB và tam giác MBH ta có
^M _ chung
^MIB = ^MBH = 900
Vậy tam giác MIB ~ tam giác MBH (g.g)
\(\dfrac{MB}{MH}=\dfrac{MI}{MB}\Rightarrow MB^2=MI.MH\)
ko bt:)
:)