Toán

4) Ta có: \(x^2\ge0\)

⇔ \(-x^2\le0\)

⇔ \(4-x^2\le4\)

⇔ \(\sqrt{4-x^2}\le2\)

Dấu = xảy ra khi \(x=2\)

Vậy GTLN của A là 2 khi x=2

5) Ta có: \(\sin^2\alpha+cos^2\alpha=1\)

⇒ \(cos^2\alpha=1-\dfrac{1}{4}\)

⇒ \(cos^2\alpha=\dfrac{3}{4}\)

⇒ \(\cos\alpha=\dfrac{\sqrt{3}}{2}\)

Ta có: \(\tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)

\(tan\alpha=\dfrac{1}{2}:\dfrac{\sqrt{3}}{2}\)

\(tan\alpha=\dfrac{\sqrt{3}}{3}\)

Ta có: \(cot\alpha=\dfrac{1}{tan\alpha}\)

\(cot\alpha=\dfrac{3}{\sqrt{3}}\)

\(cot\alpha=\sqrt{3}\)

\(=\dfrac{1993\left(1994-1992\right)}{1993\cdot1992+1993\cdot7+1993}\)

\(=\dfrac{2}{1992+7+1}=\dfrac{2}{2000}=\dfrac{1}{1000}\)

=9/96+44/96+10/96

=63/96

=21/32

\(=7^{98}\left(7^2+7-1\right)=7^{98}\cdot55⋮55\)

`27^400 = 3^1200`

`81^525 = 3^2020`

`-> 27^400 < 81^525`

27⁴⁰⁰ = (3³)⁴⁰⁰ = 3¹²⁰⁰

81⁵²⁵ = (3⁴)⁵²⁵ = 3²¹⁰⁰

Do 1200 < 2100 nên 3¹²⁰⁰ < 3²¹⁰⁰

Vậy 27⁴⁰⁰ < 81⁵²⁵

Diện tích viên gạch là:

`60xx60=3600(cm^2)`

`80m^2=800000cm^2`

Diện tích viên gạch là:

`60xx60=3600(cm^2)`

Số viên gạch cần dùng là:

\(800000:3600\approx223\left(viên\right)\)

D/s....

ai giúp tui vs giải chi tiết vs

`(x-1)/3 = x + 1`

`->x - 1 = 3x + 3`

`-> -2x = 4`

`-> x = -2`

sos

=>3x+3=x-1

=>2x=-4

=>x=-2

`12 = 3 xx 4`

`1122 = 33 xx 2 xx 17`

`111222 = 2 xx 3^2 xx 37 xx 167`

`12= 3 xx 4`

`1122= 33 xx 34`

`111222=333 xx 444`

`(x- 1/2) .1/3 +5/7 = 9 5/7`

`=> (x- 1/2) .1/3 +5/7 =68/7`

`=>  (x- 1/2) .1/3=68/7-5/7`

`=>  (x- 1/2) .1/3=9`

`=>x-1/2 = 9 : 1/3`

`=>x-1/2 = 9 .3`

`=>x-1/2 =27`

`=>x=27 + 1/2`

`=>x= 54/2 + 1/2`

`=>x= 55/2`

`(x-1/2) . 1/3 + 5/7 = 9 5/7`

`(x-1/2) . 1/3 + 5/7 =68/7`

`(x-1/2) . 1/3 = 68/7 - 5/7`

`(x-1/2) .1/3= 63/7`

`(x-1/2) . 1/3 = 9`

`x-1/2 = 9 : 1/3`

`x-1/2 = 9 .3`

`x-1/2 = 27`

`x=27+1/2`

`x=54/2 + 1/2`

`x=55/2`

Vậy `x=55/2`

2: \(\dfrac{12}{\sqrt{2}}=6\sqrt{2}=\sqrt{72}< \sqrt{125}=\sqrt{5}\cdot5\)

4: \(\sqrt{\sqrt{2}}\cdot\left(\sqrt{3}+1\right)=\sqrt{\sqrt{2}\left(4+2\sqrt{3}\right)}\)

\(=\sqrt{4\sqrt{2}+2\sqrt{6}}\)