1 The gymnast won a total of ten medals at 3 Olympic Games

2 The principal invited a sports star to give a talk at my school yesterday

\(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\left(đk:x\ge-1\right)\)

\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)

\(\Leftrightarrow2\sqrt{x+1}=4\)

\(\Leftrightarrow\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\left(tm\right)\)

Bài 2:

e) \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=\sqrt{x-2}-12\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}-12.\sqrt{\dfrac{1}{9}}.\sqrt{x-2}=\sqrt{x-2}-12\)

\(\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=\sqrt{x-2}-12\)

\(\Leftrightarrow3\sqrt{x-2}=12\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\Leftrightarrow x=18\left(tm\right)\)

d) \(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\left(đk:x\ge-1\right)\)

\(\Leftrightarrow\sqrt{9}.\sqrt{x+1}-2.\sqrt{\dfrac{1}{4}}.\sqrt{x+1}=4\)

\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)

\(\Leftrightarrow2\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=2\)

\(\Leftrightarrow x+1=4\Leftrightarrow x=3\)

Bài 2:

e) \(\sqrt{4x-8}-12.\sqrt{\dfrac{x-2}{9}}=\sqrt{x-2}-12\Rightarrow\sqrt{4}.\sqrt{x-2}-12.\sqrt{\dfrac{1}{9}}.\sqrt{x-2}=\sqrt{x-2}-12\Rightarrow2\sqrt{x-2}-4.\sqrt{x-2}=\sqrt{x-2}-12\Rightarrow3.\sqrt{x-2}=12\Rightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Rightarrow x=18\left(tm\right)\)

c: Ta có: AM//BC

AE⊥BC

Do đó:AM⊥AE

Suy ra: \(\widehat{AME}+\widehat{AEM}=90^0\)

hay \(\widehat{AME}+\widehat{BAD}=90^0\)

a: \(268_{10}=10C_{16}\)

b: \(12C_{16}=\text{100101100}_2\)

`a)P(x)+Q(x)=x^5-2x^2+1`

`=>Q(x)=x^5-2x^2+1-P(x)`

`=>Q(x)=x^5-2x^2+1-x^4+3x^2-1/2+x`

`=>Q(x)=x^5-x^4+x^2+x+1/2`

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`b)P(x)-R(x)=x^3`

`=>R(x)=P(x)-x^3`

`=>R(x)=x^4-3x^2+1/2-x-x^3`

`=>R(x)=x^4-x^3-3x^2-x+1/2`

Ta có:

\(P\left(x\right)+Q\left(x\right)=x^5-2x^2+1\)

\(\Rightarrow Q\left(x\right)=P\left(x\right)-\left(x^5-2x^2+1\right)\)

\(=x^4-3x^2+\dfrac{1}{2}-x-x^5+2x^2-1\)

\(=-x^5+x^4-x^2-x-\dfrac{1}{2}\)

Vậy \(Q\left(x\right)=-5^2+x^4-x^2-x-\dfrac{1}{2}\)

a) <=> Q(x) = (x⁵ - 2x² + 1) - P(x)

= (x⁵ - 2x² + 1) - (x⁴ - 3x² + 1/2 - x)

= x⁵ - 2x² + 1 - x⁴ + 3x² + x - 1/2

= x⁵ - x⁴ + x² + x + 1/2

Vậy Q(x) = x⁵ - x⁴ + x² + x + 1/2