Toán

Các bạn giúp mình với mình đang cần gấp ạ. Bạn nào giải được thì giải giúp mình trước 3h với ạ ! Mình cảm ơn 

\(5^3\cdot5^{x-2}=25^2\)

=>5^(x+1)=5^4

=>x+1=4

=>x=3

5^3 * 5x -2 = 25^2

5^4 * x-2 = (5^2)^2

5^4 * x - 2 = 5^4

x - 2/5^4 = 1

x - 2/625 = 1

x = 627/625

\(134-\left(y-3\right)=35+160:5\)

\(\Rightarrow134-\left(y-3\right)=35+32\)

\(\Rightarrow134-\left(y-3\right)=67\)

\(\Rightarrow y-3=134-67\)

\(\Rightarrow y-3=67\)

\(\Rightarrow y=67+3\)

\(\Rightarrow y=70\)

`134-(y-3)=35+160:5`

`=>134-(y-3)=35+32`

`=>y-3=134-(35+32)`

`=>y-3=134-67`

`=>y-3=67`

`=>y=70`

Vậy `x\in{70}`

`1. P = x/(sqrt x-1)`

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

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

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

`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.

ĐTXR `<=> (sqrtx-1)^2 = 1`.

`<=> x =4` hoặc `x = 0 ( ktm)`.

Vậy Min A `= 4 <=> x= 4`.

1) \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{(x-\sqrt{x})+(\sqrt{x}-1)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}+1\)

\(=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)

Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)

Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)

\(\Rightarrow P\ge2+2=4\)

Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

KL;....

2:

\(B=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}\)

\(=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

=>\(B>=2\cdot\sqrt{25}-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{5^2-2\cdot5\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)

\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{25}\)

\(=\sqrt{5^2}\)

\(=5\)

\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}=\sqrt{25}=5\)

Bài 1: `m = 3 + 3^2 + ... + 3^2019`.

`= 3(1+3+3^2) + 3^4(1+3+3^2) + 3^7(1+3+3^2) + ... + 3^2017(1+3+3^2)`

`= 13 . 3 + 13 . 3^4 + 3^7 . 13 + ... + 3^2017 .13`

`= 13(3+3^4+3^7+...+3^2017) vdots 13`.

Bài 4: `(n+2021^2022)(n+2022^2021)`

`@ n cancel vdots 2 => n + 2021^2022 equiv 1 + 1 equiv 2 equiv 0 ( mod 2)`

`@ n vdots 2 <=> n + 2022^2021 equiv 0 + 0 equiv 0 ( mod 2)`

Vậy ta có dpcm

3:

\(A=\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2020}\left(5+5^2+5^3+5^4\right)\)

=780(1+5^4+...+5^2020) chia hết cho 65

2:

\(A=3\left(1+3^2+3^4\right)+...+3^{2011}\left(1+3^2+3^4\right)\)

=91(3+...+3^2011) chia hết cho 13

A=3(1+3^2+3^4+3^6)+...+3^2009(1+3^2+3^4+3^6)

=820*(3+...+3^2009) chia hết cho 41

Số lít dầu ở thùng to là:

\(\left(6000+1800\right):2=3900\left(l\right)\)

Số lít dầu ở thùng nhỏ là:

\(3900-1800=2100\left(l\right)\)

Đáp số: ...

Số lít dầu ở thùng to là:

\(\left(6000+1800\right):2=3900\left(l\right)\)

Số lít dầu ở thùng nhỏ:

3900-1800=2100 (l)

Đáp số: Thùng to: 3900 (l)

             Thùng nhỏ: 2100(l)

\(45\cdot y-30=12\cdot15-40\cdot3\)

\(\Rightarrow45\cdot y-30=180-120\)

\(\Rightarrow45\cdot y-30=60\)

\(\Rightarrow45\cdot y=60+30\)

\(\Rightarrow45\cdot y=90\)

\(\Rightarrow y=90:45\)

\(\Rightarrow y=2\)

\(\Leftrightarrow45y-30=60\Leftrightarrow45y=90\Leftrightarrow y=2\)

Bài 2: ước của:

`12: +-1; +-2;+-3;+-4; +-6;+-12`

`28: +-1; +-2;+-4:+-7:+-14;+-28`

`47: +-1; +-47`

`105: +-1; +-3; +-5; +-7;+-15;+-21;+-35; +-105`.

`15:+-1;+-3;+-5;+-15`

`32: +-1;+-2;+-4;+-8;+-16;+-32`

`66: +-1;+-2;+-3; +-6;+-11;+-22;+-33;+-66`

`102: +-1;+-2;+-3;+-6;+-17; +-34; +-51; +-102`.

Bài 4: `22 = 2 xx 11`

`34 = 2 xx 17`

`73 = 1 xx 73`

`99 = 9 xx 11 = 3 xx 3 xx 11`

Bài 7: `16: +-1;+-2;+-4:+-8;+-16`

`37: +-1;+-37`

`46: +-1; +-2; +-23;+-46`

`49: +-1; +-7; +-49`.

\(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\dfrac{\sqrt{15-10\sqrt{2}}-\sqrt{10}}{\sqrt{5}}\)

\(=\left(\dfrac{-\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}-2\right)\left(\dfrac{\sqrt{5}\left(3+\sqrt{5}\right)}{3+\sqrt{5}}-2\right)\dfrac{\sqrt{5}.(\sqrt{3-2\sqrt{2}}-\sqrt{2})}{\sqrt{5}}\)

\(=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right).(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{2})\)

\(=-3.-1=3\)

`((5-2 sqrt 5)/(2-sqrt5) - 2)((5+3 sqrt 5)/(3+sqrt 5) - 2)`

`= (5-2sqrt 5 - 4 + 2 sqrt 5)/(2 -sqrt 5) . (5+3 sqrt 5 - 6 - 2 sqrt 5)/(3 + sqrt 5)`

`= 1/(2-sqrt5) . (-1 + sqrt 5)/(3 + sqrt 5)`

`= (sqrt 5 - 1)/((sqrt 5 + 3)(2 - sqrt 5))`

`b, (sqrt(15-10sqrt2) - sqrt 10)/(sqrt 5)`

`= (sqrt(10 - 2 . sqrt 5.sqrt 10 + 5) - sqrt 10)/(sqrt 5)`

`= (sqrt 10 - sqrt 5 - sqrt 10)/(sqrt5)`

`= -1`

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)

`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`