a)

(x-2)²\(\ge\))

(y-3)²\(\ge\)0

=> (x-2)²=(y-3)²=0

=>\(\begin{cases}x-2=0\\y-3=0\end{cases}\Rightarrowy=3}}\)

b)

để 5^(x-2)(x+3)=1

=> (x-2)(x+3)=0

=> \(\left[\begin{array}{nghiempt}x-2=0\\x+3=0\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.}\)

a)\(\left(x+2\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\begin{cases}x+2=0\\y-3=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-2\\y=3\end{cases}\)

Vậy x=-2 ; y=3

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

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=3\end{array}\right.\)

Ta có : 3x - 7/3 - 2x - 1/2 = 7 .

=>       x ( 3 - 2 ) - ( 7/3 + 1/2 ) = 7 .

=>       x - ( 14/6 + 3/6 ) = 7 .

=>            x - 17/6         = 7 .

=>                x                = 7 + 17/6 .

=>                x                 = 59/6 .

vậy x = 59/6 .

\(3x-\frac{7}{3}-2x-\frac{1}{2}=7\)

\(\Leftrightarrow\left(3x-2x\right)-\left(\frac{7}{3}+\frac{1}{2}\right)=7\)

\(\Leftrightarrow x-\frac{17}{6}=7\)

\(\Leftrightarrow x=7+\frac{17}{6}\)

\(\Leftrightarrow x=\frac{59}{6}\)

\(\frac{3x-7}{3}\)  -    \(\frac{2x-1}{2}\) \(=\) \(7\)

\(\frac{3x-7}{3}\) = \(7\) +  \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  = \(\frac{14}{2}\) +   \(\frac{2x-1}{2}\)

\(\frac{3x-7}{3}\)  =   \(\frac{2x+13}{2}\)

<=> \(\left(3x-7\right)\)  x  \(2\) =    \(\left(2x+13\right)\)x   \(3\)

<=> \(6x-14=6x+39\) <=> 6x - 6x = 14 + 39

                                                       <=> 0x = 53 => x ko tốn tại

vậy x ko tồn tại

Đề bài so sánh thì k có chữ li nhé mk đánh nhâm

Câu 2:

a: \(\left(x-1\right)^3=27\)

=>x-1=3

=>x=4

b: \(\Leftrightarrow2x=25-1=24\)

=>x=12

c: \(\left(2x-1\right)^3=-8\)

=>2x-1=-2

=>2x=-1

=>x=-1/2

d: \(\left(x-1\right)^{1999}=\left(x-1\right)^{2002}\)

\(\Leftrightarrow\left(x-1\right)^{1999}\left[\left(x-1\right)^3-1\right]=0\)

=>(x-1)(x-2)=0

=>x=1 hoặc x=2

Đề bài có phải là như vậy ko bạn:

5^x-2 - 3²= 2⁴- (6⁸ : 6⁶ - 6²) ( x^2-1)⁴ =81

( 3^x+4 )² = 19⁶ : ( 19³ x 19² ) - 3^{1x 2005}

Bạn xem hộ mik như vậy có đúng đề ko, đúng thì mik làm cho nhé!

\(\frac{2}{2.3}\) +   \(\frac{2}{3.4}\) +  \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\) 

2 . (  \(\frac{1}{2}\) -  \(\frac{1}{3}\) + \(\frac{1}{3}\) -  \(\frac{1}{4}\) + .......+  \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)

2 . ( \(\frac{1}{2}\) -  \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)

\(\frac{1}{2}\) -  \(\frac{1}{x+1}\) =  \(\frac{2017}{2019}\) : 2 

 \(\frac{1}{2}\) -  \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)

             \(\frac{1}{x+1}\)  =  \(\frac{1}{2}\)  -    \(\frac{2017}{4038}\)

              \(\frac{1}{x+1}\)  = \(\frac{1}{2019}\) 

     <=> x + 1 = 2019 => x = 2018

vậy x = 2018

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)

\(\Rightarrow x+1=2019\)

\(\Leftrightarrow x=2018\)

Vậy  \(x=2018\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2017}{2019}:2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2017}{4038}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)

=> x + 1 = 2019

=> x = 2018

Xét \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=\frac{x_2^3-x_1^3}{x_2-x_1}=\frac{\left(x_2-x_1\right)\left(x_2^2+x_1x_2+x_1^2\right)}{x_2-x_1}=x_1^2+x_1x_2+x_2^2=\left(x_1^2+x_1x_2+\frac{x_2^2}{4}\right)+\frac{3x_2^2}{4}\)

\(=\left(x_1+\frac{x_2}{2}\right)^2+\frac{3x_2^2}{4}>0\)

Do vậy hàm số luôn đồng biến.

Với x₁ > x₂ thì f(x​₁) - f(x₂)

= x₁³ - x₂³ = (x₁ - x₂)(x₁² + x₁ x₂ + x₂²) = (x₁ - x₂)[(x₁² + x₁ x₂ + x₂²/4) + 3x₂² ) = (x₁ - x₂)[x₁ + x₂/2)² + 3x₂²/4) > 0

Vậy hàm số đồng biến

a. `4x^2-20x+25=0`

`<=>(2x)^2-2.2x.5 +5^2=0`

`<=>(2x-5)^2=0`

`<=>2x-5=0`

`<=>x=5/2`

b. `(x-5)(x+5)-(x-3)^2=2(x-7)`

`<=>x^2-25-x^2+6x-9=2x-14`

`<=>6x-34=2x-14`

`<=>4x=20`

`<=>x=5`

\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)

\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)

\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)

a) Có: (2x)² - 2.2.5.x + 5² = 0

          ⇒ (2x - 5)² = 0 ⇒ 2x - 5 = 0

          ⇒ 2x = 5 ⇒ x = \(\dfrac{5}{2}\)

b) Có: x² - 25 - x² + 6x - 9 = 2x - 14

           ⇒ 6x - 36 = 2x - 14

           ⇒ 4x = 22

           ⇒ x = \(\dfrac{11}{2}\)

\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)

<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)

<=>  \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)

<=> \(\frac{1}{x+2}=\frac{1}{18}\)

=>  \(x+2=18\)

<=>  \(x=16\)

Vậy...