2x+3x=5x

5x=1505

 x =1505:5

 x = 301

1. Tính :

\(\left(1+3+5+7+...+2017\right)\cdot\left(135135\cdot137-135\cdot137137\right)\)

= (1 + 3 + 5 + 7 + ... + 2017) . (135.1001.137 - 135.137.1001)

= (1 + 3 + 5 + 7 + ... + 2017).0 = 0

2. Tìm x:

a) 2x  + 3x =  1505 => 5x = 1505 => x = 301

b) 1 + 3 + 5 + ... + x = 3200 (sửa lại cái đề đây nhé)

Số số hạng là : (x - 1) : 2 + 1 = \(\frac{x+1}{2}\)

Tổng : \(\left(1+x\right)\cdot\frac{x+1}{2}=\frac{\left(x+1\right)\left(x+1\right)}{2}=\frac{\left(x+1\right)^2}{2}\)

=> \(\frac{\left(x+1\right)^2}{2}=3200\)

=> \(\left(x+1\right)^2=6400\)

=> \(\left(x+1\right)^2=\left(\pm80\right)^2\)

=> \(\orbr{\begin{cases}x+1=80\\x+1=-80\end{cases}}\Rightarrow\orbr{\begin{cases}x=79\\x=-81\end{cases}}\)

x = -81 loại vì x thuộc số tự nhiên => x = 79

c) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 5750

=> x + 1 + x + 2 + x + 3 + ... + x + 100 = 5750

=> (x + x  +... +x) + (1 + 2 + 3 + ... +100) = 5750

Số số hạng : (100 - 1) : 1 + 1 = 100(số)

Tổng : (1 + 100).100 : 2  = 5050

=> 100x + 5050 = 5750

=> 100x = 700

=> x = 7

3. a)

A = 123.123

B = 121.124 = (123 - 2)(123 + 1) = 123² - 1² = 123.123 - 1

=> A > B

b) C = 123.137137 = 123.1001.137

D = 137.123123 = 137.1001.123

=> C = D

c) E = 2015.2017 = 2015.(2016 + 1) = 2015.2016 + 2015 (1)

F = 2016.2016 = (2015 + 1).2016 = 2015.2016 + 2016(2)

Từ (1) và (2) => E < F (vì 2015 < 2016)

a. \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)

\(\left(\frac{1}{3}x\right):\frac{2}{3}=\frac{35}{8}\)

\(\frac{1}{3}x=\frac{35}{8}.\frac{2}{3}=\frac{35}{12}\)

\(x=\frac{35}{12}.3=\frac{35}{4}\)

b) 4,5:0,3=2,25:(0,1.x)

\(\frac{45}{10}:\frac{10}{3}=\frac{225}{100}:\left(\frac{1}{10}x\right)\)

\(15=\frac{9}{4}:\left(\frac{1}{10}x\right)\)

\(\frac{9}{4}:\left(\frac{1}{10}x\right)=15\)

\(\frac{1}{10}x=\frac{9}{4}.\frac{1}{15}=\frac{3}{20}\)

\(x=\frac{3}{20}.10=\frac{3}{2}\)

c) 8:\(\left(\frac{1}{4}x\right)=2:0,02\)

8:\(\left(\frac{1}{4}x\right)\)=2:\(\frac{2}{100}\)

8:\(\left(\frac{1}{4}x\right)=2.\frac{100}{2}=100\)

\(\frac{1}{4}x=\frac{8}{100}\)

\(x=\frac{8}{100}.4=\frac{8}{25}\)

d)\(3:2\frac{1}{4}=\frac{3}{4}:\left(6x\right)\)

\(3.\frac{9}{4}=\frac{3}{4}:\left(6x\right)\)

\(\frac{3}{4}:\left(6x\right)=\frac{27}{4}\)

\(6x=\frac{3}{4}.\frac{4}{27}=\frac{1}{9}\)

\(x=\frac{1}{9}.\frac{1}{6}=\frac{1}{54}\)

 a.=> 1/2 . x = 35/8

 => x = 31/8

 b. => 15 = 22,5 . x

=> x = 2/3

c. => 32 . x =  100

=>x=25/8 

d. 4/3= 1/8 . x 

x = 32/3

a) 

\(2x>2\left(x+1\right)\) 

\(2x>2x+2\) 

\(0>2\left(sai\right)\) 

Vậy bất phương trình vô nghiệm   

\(\frac{3x-5}{x+2}< 4\) 

\(3x-5< 4\left(x+2\right)\) 

\(3x-5< 4x+8\) 

\(-13< x\) 

Vậy x > -13 thỏa bất phương trình 

2x/x+1 = 2x+2-2/x+1 = 2 - 2/x+1 >2 => 2/x+1<0 => x<-1
3x-5/x+2  - 3 = 3x+6-11/x+2   -  3  = 3   -   11/x+2  -  3 = -11/x+2 <1 => x>-2

Câu a đề hơi sai nha bạn, nên mình chỉ giải câu b thoi

Áp dụng AM-GM cho các bộ 3 số dương (x,y,z) và (1/x,1/y,1/z):

\(x+y+z\ge3\sqrt[3]{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\)

\(\Rightarrow P\ge6\sqrt[3]{xyz}+\frac{3}{\sqrt[3]{xyz}}\ge2\sqrt{6\sqrt[3]{xyz}.\frac{3}{\sqrt[3]{xyz}}}=6\sqrt{2}\)(BĐT Cô-si)

Dấu = xảy ra khi và chỉ khi \(x=y=z=\frac{1}{\sqrt{2}}\)( thỏa x,y,z thuộc (0;1))

Mình cần câu a ạ :<

a) 

Vì \(x^2-2x+3=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)  

\(\Rightarrow x-8< 0\) 

\(x< 8\) 

b) 

Ta có : 

\(3x^2+5\ge5\forall x\)           

\(\Rightarrow7x+9>0\) 

\(7x>-9\) 

\(x>-\frac{9}{7}\)

a)\(\frac{x-8}{x^2-2x+3}< 0\)

Vì x² - 2x + 3 = ( x² - 2x + 1 ) + 2 = ( x - 1 )² + 2 ≥ 2 > 0 ∀ x

nên ta chỉ cần xét x - 8 < 0

x - 8 < 0 => x < 8

Vậy với x < 8 thì \(\frac{x-8}{x^2-2x+3}< 0\)

b)\(\frac{7x+9}{3x^2+5}>0\)

Vì 3x² + 5 ≥ 5 > 0 ∀ x

nên ta chỉ cần xét 7x + 9 > 0

7x + 9 > 0 => 7x > -9 => x > -9/7

Vậy với x > -9/7 thì \(\frac{7x+9}{3x^2+5}>0\)

+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)    \(\left(ĐK:x\ge0\right)\)

        \(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)

        \(\Leftrightarrow3\sqrt{3x}=6\)

        \(\Leftrightarrow\sqrt{3x}=2\)

        \(\Leftrightarrow3x=4\)

        \(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{4}{3}\right\}\)

+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\)    \(\left(ĐK:x\ge1\right)\)

        \(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

        \(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)

        \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

        \(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)

 Vậy \(S=\left\{1,15\right\}\)

+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\)       \(\left(ĐK:x\ge0\right)\)

         \(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

         \(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

         \(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

   Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)

    \(\Rightarrow\)\(\sqrt{x}-4< 0\)

   \(\Leftrightarrow\)\(\sqrt{x}< 4\)

   \(\Leftrightarrow\)\(x< 16\)

   Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)

 Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)

\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\)  (Đk: x \(\ge\)0)

<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)

<=> \(3\sqrt{3x}=6\)

<=> \(\sqrt{3x}=2\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)

<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)

<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)

<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\) 

<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)

\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)

<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)

<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)

<=>  \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)

Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)

<=> \(\sqrt{x}< 4\) <=> \(x< 16\)

Kết hợp với đk => S = {x|0 < x < 16}

\(a,4\left(2-x\right)^2+xy-2y\)

\(=4\left(2-x\right)^2-y\left(2-x\right)\)

\(=4-y\left(2-x\right)^2\left(2-x\right)\)

\(=\left(2-x\right)\left[\left(2-x\right)4-y\right]\)

\(=\left(2-x\right)\left(4x-8+y\right)\)

\(c,x^3+y^3+z^3-3xyz\)

\(=x^3+y^3+z^3+3x^2y-3x^2y+3xy^2-3xy^2-3xyz\)

\(=\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+1\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz\)

\(=\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

a) 4(2 - x)² + xy - 2y = 4(x - 2)² + y(x - 2) = (4x - 8 + y)(x - 2)

b) 2(x - 1)³ - 5(x - 1)² - (x - 1) = (x - 1)[2(x - 1)² - 5(x - 1) - 1]

= (x - 1)(2x² - 4x + 2 - 5x + 5 - 1) = (x - 1)(2x² - 9x + 6)

c) x³ + y³ + z³ - 3xyz = (x + y)(x² - xy + y²) + z³ - 3xyz

= (x + y)³ + z³ - 3xy(x + y) - 3xyz = (x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z)

= (x + y + z)(x² + y² + z² - xz - yz + 2xy - 3xy) = (x + y + z)(x² + y² + z² - xy - yz - xz)

+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)

         \(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)

         \(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)

         \(=4\sqrt{3}\approx6,9282\)

+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)

        \(=\sqrt{x-9+6\sqrt{x-9}+9}\)

        \(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)

        \(=\left|\sqrt{x-9}-3\right|\)

\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)

x² + y² = 2x²y²

<=> 2x² + 2y² - 4x²y² = 0

<=> 2x²(1 - 2y²) - (1 - 2y²) = -1

<=> (2x² - 1)(2y² - 1) = 1 = 1.1

Lập bảng: 

Vậy ...

OK cảm ơn nha