Toán

Ta có:

\(2^{3000}=\left(2^3\right)^{1000}=8^{1000}\)

\(3^{2000}=\left(3^2\right)^{1000}=9^{1000}\)

Mà: \(8< 9\Rightarrow8^{1000}< 9^{1000}\)

\(\Rightarrow2^{3000}< 3^{2000}\)

`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

=(x+1)^2-2(2x-1)(x+1)+(2x-1)^2

=(x+1-2x+1)^2

=(-x+2)^2=x^2-4x+4

9. a) \(\dfrac{4}{3}+\dfrac{2}{3}:x=-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{26}{15}\)\(\Rightarrow x=\dfrac{2}{3}:\left(-\dfrac{26}{15}\right)=-\dfrac{5}{13}\)

b) \(\dfrac{7}{3}-1\dfrac{3}{5}.x=-1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{8}{5}x=\dfrac{7}{3}-\left(-\dfrac{5}{3}\right)=\dfrac{7+5}{3}\)

\(\Rightarrow\dfrac{8}{5}x=4\Leftrightarrow x=\dfrac{5}{2}\)

c) \(\left(-2\dfrac{1}{4}:x+1,5\right)\left(\dfrac{-5}{3}.x-\dfrac{5}{12}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-2\dfrac{1}{4}:x=-1.5\\\dfrac{-5}{3}.x=\dfrac{5}{12}\end{matrix}\right.\)         \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

d) \(-\dfrac{3}{5}.x-\left(x+2,5\right)=-\dfrac{13}{15}\)

\(\Rightarrow-\dfrac{8}{5}.x=-\dfrac{13}{15}+2,5\)

\(\Rightarrow-\dfrac{8}{5}.x=\dfrac{49}{30}\)

\(\Rightarrow x=\dfrac{49}{30}:\left(\dfrac{-8}{5}\right)=-\dfrac{49}{48}\)

10.

a) \(\left(-3+\dfrac{3}{x}-\dfrac{1}{3}\right):\left(1+\dfrac{2}{5}+\dfrac{2}{3}\right)=-\dfrac{5}{4}\)

\(\Rightarrow\left(\dfrac{3}{x}-\dfrac{10}{3}\right):\dfrac{31}{15}=-\dfrac{5}{4}\)

\(\Rightarrow\dfrac{3}{x}-\dfrac{10}{3}=-\dfrac{5}{4}.\dfrac{31}{15}=-\dfrac{31}{12}\)

\(\Rightarrow\dfrac{3}{x}=-\dfrac{31}{12}+\dfrac{10}{3}=\dfrac{3}{4}\)

\(\Leftrightarrow x=4\)

b) \(\left(\dfrac{2}{5}-x\right):1\dfrac{1}{3}+\dfrac{1}{2}=-4\)

\(\Rightarrow\left(\dfrac{2}{5}-x\right):\dfrac{4}{3}=-4-\dfrac{1}{2}=-\dfrac{9}{2}\)

\(\Rightarrow\dfrac{2}{5}-x=-\dfrac{9}{2}.\dfrac{4}{3}=-6\)

\(\Rightarrow x=\dfrac{2}{5}-\left(-6\right)=\dfrac{32}{5}\)

c) \(-\dfrac{3x}{4}.\left(\dfrac{1}{x}+\dfrac{2}{7}\right)=0\)  

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3x}{4}=0\\\dfrac{1}{x}+\dfrac{2}{7}=0\end{matrix}\right.\)                 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{x}=-\dfrac{2}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{2}\end{matrix}\right.\)

d) \(2,5x+\dfrac{-4}{7}=\dfrac{1}{2}x\)

\(\Rightarrow2,5x-\dfrac{1}{2}x=\dfrac{4}{7}\)

\(\Rightarrow2x=\dfrac{4}{7}\Leftrightarrow x=\dfrac{2}{7}\)

a) 8⁴ = (2³)⁴ = 2¹²

b) (2⁴)³ = 2¹²

c) 16³ = (2⁴)³ = 2¹²

d) 27³ = (3³)³ = 3⁹

a) \(8^4\)

\(=\left(2^3\right)^4\)

\(=2^{3\cdot4}\)

\(=2^{12}\)

b) \(\left(2^4\right)^3\)

\(=2^{4\cdot3}\)

\(=2^{12}\)

c) \(16^3\)

\(=\left(2^4\right)^3\)

\(=2^{4\cdot3}\)

\(=2^{12}\)

d) \(27^3\)

\(=\left(3^3\right)^3\)

\(=3^{3\cdot3}\)

\(=3^9\)

a: Xét ΔCBD có
CA vừa là đường cao, vừa là đường trung tuyến

nên ΔCBD cân tại C

c: Gọi N là trung điểm của AC

=>QN là đường trung trực của AC

=>QN//AD

Xét ΔCAD có

N là trung điểm của AC

NQ//AD

=>Q là trung điểm của CD

Xét ΔCDB có

CA,DK là trung tuyến
CA cắt DK tại M

=>M là trọng tâm

mà BQ là trung tuyến

nên B,M,Q thẳng hàng

`\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}`

`=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}`

`=\sqrt{(3-\sqrt{6})^2}+\sqrt{(2\sqrt{6}-3)^2}`

`=3-\sqrt{6}+2\sqrt{6}-3`

`=\sqrt{6}`

hum

\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-\sqrt{6^2\cdot6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\) 

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot3\cdot\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}\)

\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)

\(=-\sqrt{6}+3+2\sqrt{6}-3\)

\(=\sqrt{6}\)

a) ĐKXĐ: \(x\ne4,x\ge0\)

b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{x+\sqrt{x}+2\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

c) \(P=2\) khi:

\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}=4\)

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

\(\Leftrightarrow x=16\left(tm\right)\)

d) P nguên khi:

\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6-6}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)-6}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\)

Phải nguyên:

\(\Rightarrow6\) ⋮ \(\sqrt{x}+2\) 

\(\Rightarrow\sqrt{x}+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Mà: \(x\ge0\) và \(x\ne4\) nên

\(\sqrt{x}+2\in\left\{2;3;6\right\}\)

\(\Rightarrow x\in\left\{0;1;16\right\}\)

giai cau d th ah

`a)P` xác định `<=>{(x >= 0),(x ne 4):}`

`b)` Với `x >= 0,x ne 4` có:

`P=[\sqrt{x}+1]/[\sqrt{x}-2]+[2\sqrt{x}]/[\sqrt{x}+2]+[2+5\sqrt{x}]/[4-x]`

`P=[(\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-2-5\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`P=[x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`

`P=[3x-6\sqrt{x}]/[(\sqrt{x}+2)(\sqrt{x}-2)]`

`P=[3\sqrt{x}(\sqrt{x}-2)]/[(\sqrt{x}+2)(\sqrt{x}-2)]=[3\sqrt{x}]/[\sqrt{x}+2]`

`c)` Với `x >= 0,x ne 4` có:

`P=2<=>[3\sqrt{x}]/[\sqrt{x}+2]=2`

    `<=>3\sqrt{x}=2\sqrt{x}+4`

   `<=>\sqrt{x}=4<=>x=16` (t/m)

`d)` Với `x >= 0,x ne 4` có:

`P=[3\sqrt{x}]/[\sqrt{x}+2]=[3\sqrt{x}+6-6]/[\sqrt{x}+2]=3-6/[\sqrt{x}+2]`

`P` nguyên `<=>3-6/[\sqrt{x}+2] in Z`

     `=>6/[\sqrt{x}+2] in ZZ`

 `=>\sqrt{x}+2 in Ư_6`

Mà `Ư_6 ={+-1;+-2;+-3;+-6}`

Ta có bảng:

\begin{array}{|c|c|c|}\hline \sqrt{x}+2&1&-1&2&-2&3&-3&6&-6\\\hline x&L&L&0&L&1&L&16&L \\\hline\end{array}

  Mà `x >= 0,x ne 4, x in Z =>x={0;1;16}`

`a)` Hàm số xác định `<=>{(5x+3 >= 0),(2x+1 >= 0):}`

                                 `<=>{(x >= -3/5),(x >= -1/2):}<=>x >= -1/2`

`b)` Hàm số xác định `<=>{(x-7 >= 0),(14-x >= 0):}`

                    `<=>{(x >= 7),(x <= 14):}<=>7 <= x <= 14`

`@` `\text {Ans}`

`\downarrow`

`16,`

`a)`

`7^6 \div 7^4 + 3^3 * 3^2 - 1^(2022)`

`= 7^2 + 3^5 - 1`

`= 49 + 243 - 1`

`=291`

`b)`

`6^6 \div 6^5 + 2^3 * 2^2 - 2022^0`

`= 6 + 2^5 - 1`

`= 6 + 32 - 1`

`=37`

`c)`

`2^3 * 2^2 + 6^6 \div 6^5 + 23^1`

`= 2^5 + 6 + 23`

`= 32 + 6 + 23`

`=61`

`d)`

`3^3 * 3^2 + 7^7 \div 7^5 - 2^3` dk bạn;-;?

`= 3^5 + 7^2 - 8`

`= 243 + 49 - 8`

`=284`

a) \(7^6:7^4+3^3\cdot3^2-1^{2022}\)

\(=7^2+3^5-1\)

\(=49+243-1\)

\(=49+242\)

\(=291\)

b) \(6^6:6^5+2^3\cdot2^2-2022^0\)

\(=6+2^5-1\)

\(=6+32-1\)

\(=32+5\)

\(=37\)

c) \(2^3\cdot2^2+6^6:6^5+23^1\)

\(=2^5+6+23\)

\(=32+29\)

\(=61\)

d) \(3^3\cdot3^2+7^7:7^5-2^3\)

\(=3^5+7^2-8\)

\(=243+49-8\)

\(=243+41\)

\(=284\)

a) 7⁶ : 7⁴ + 3³.3² - 1²⁰²²

 = 7² + 3⁵ - 1

= 49 + 242  = 291

b) 6⁶ : 6⁵ + 2³.2² - 2022⁰

= 6 + 2⁵ - 1

= 5 + 32 = 37

c) 2³.2² + 6⁸:6⁵ + 23¹

= 2⁵ + 6³ + 23

= 32 + 216 + 23

= 271

d) 3³.3² + 7⁷:7⁵ - 2³

= 3⁵ + 7² - 8

= 243 + 49 - 8

= 284