Để A đạt GTLN

=>x² -2x đạt giá trị dương nhỏ nhất

=>x²-2x=1

=>x²-2x-1=0

=>x=$1-\sqrt{2};\sqrt{2}+1$1−√2;√2+1

Vậy A ko xảy ra GTLN

Để A đạt GTLN

=>x² -2x đạt giá trị dương nhỏ nhất

=>x²-2x=1

=>x²-2x-1=0

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

Vậy A ko xảy ra GTLN

\(A=\frac{1}{x^2-2x}\)

A đạt giá trị lớn nhất \(\Leftrightarrow\)\(x^2-2x\)đạt giá trị nhỏ nhất.

Xét : \(B=x^2-2x=\left(x^2-2x+1\right)-1=\left(x-1\right)^2-1\ge-1\)

\(\Rightarrow MinB=-1\Leftrightarrow x=1\)

Vậy \(MaxA=-1\Leftrightarrow x=1\)

Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

\(\Rightarrow A_{min}=\dfrac{5}{2}\)

a, x+35= 2x+15+(13-4)

\(\Leftrightarrow x-2x=15-35+9\)

\(\Leftrightarrow-x=-11\)

\(\Leftrightarrow x=11\)

b, 5-2x =15-(-9)-3x

\(\Leftrightarrow-2x+3x=15+9-5\)

\(\Leftrightarrow x=19\)

c, x+135 =3x -15

\(\Leftrightarrow x-3x=-15-135\)

\(\Leftrightarrow-2x=-150\)

\(\Leftrightarrow x=\frac{-150}{-2}=75\)

a) x + 35 = 2x + 15 + ( 13 - 4 )

x - 2x = - 35 + 15 + 13 - 4

-x = - 20 + 9

-x = - 11

\(\Rightarrow\) x = 11

Vậy x = 11

b) 5 - 2x = 15 - ( -9 ) - 3x

-2x + 3x = 15 + 9 - 5

x = 24 - 5

x = 19

Vậy x = 19

c) x + 135 = 3x - 15

x - 3x = -135 - 15

-2x = -150

x = -150 : (-2)

x = 75

Vậy x = 75

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Tick cho mk nha bạn !

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Bài 2 : 

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

\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)

b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)

c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)

\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)

Bạn iair thích ý a cho mk một xíu được không bạn

a) \(\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow x=1\)

b) \(\left(2x-1\right)^3=125\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Leftrightarrow x+1=2x\)

\(\Leftrightarrow x=1\)

d) \(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

a. ( 2x + 1 )³ = 27

<=> ( 2x + 1 )³ = 3³

<=> 2x + 1 = 3

<=> 2x = 2

<=> x = 1

b. ( 2x - 1 )³ = 125

<=> ( 2x - 1 )³ = 5³

<=> 2x - 1 = 5

<=> 2x = 6

<=> x = 3

c. ( x + 1 )⁴ = 2x⁴

<=> x + 1 = 2x

<=> x = 1

d. ( 2x - 1 )⁵ = x⁵

<=> 2x - 1 = x

<=> x = 1

a) \(\left(2x+1\right)^3=27\)

\(\left(2x+1\right)^3=3^3\)

\(\Rightarrow2x+1=3\)

\(2x=3-1\)

\(2x=2\)

\(x=2\div2\)

\(x=1\)

Vậy \(x=1\).

b) \(\left(2x-1\right)^3=125\)

\(\left(2x-1\right)^3=5^3\)

\(\Rightarrow2x-1=5\)

\(2x=5+1\)

\(2x=6\)

\(x=6\div2\)

\(x=3\)

Vậy \(x=3\).

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Rightarrow x+1=2x\)

\(x-2x=-1\)

\(-x=-1\)

\(x=1\)

Vậy \(x=1\).

d) \(\left(2x-1\right)^5=x^5\)

\(\Rightarrow2x-1=x\)

\(2x-x=1\)

\(x=1\)

Vậy \(x=1\).

1: 

=>|2x+5|=5

=>2x+5=5 hoặc 2x+5=-5

=>x=0 hoặc x=-5

2: =>|x-2|=3

=>x-2=3 hoặc x-2=-3

=>x=-1 hoặc x=5

3: =>|2x-1|=1

=>2x-1=1 hoặc 2x-1=-1

=>x=0 hoặc x=1

ĐK: \(x\ne k\pi\)

\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)

\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)

\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)

\(\Leftrightarrow\left(cosx-1\right).\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

a. 6x² - (2x + 5)(3x - 2) = 7

<=> 6x² - 6x² + 4x - 15x + 10 = 7

<=> -11x = -3

<=> \(x=\dfrac{3}{11}\)

b. (5 - x)(25 + 5x + x²) + x(x² - 7) = 25

<=> 125 - x³ + x³ - 7x = 25

<=> -7x = 25 - 125

<=> -7x = -100

<=> \(x=\dfrac{100}{7}\)

c. (7 - 2x)² + (3 + 2x)(3 - 2x) = 30

<=> 49 - 28x + 4x² + 9 - 4x² = 30

<=> 4x² - 4x² - 28x = 30 - 49 - 9

<=> -28x = -28

<=> x = 1