1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

latex

Bn ơi , có phải đề bài như thế này ko?

\(5^{x+3}.5^{4x}-5^{2x}.5^{3x+2}=100\)

b, (2x) . (-4x) + 28 = 100

2x . -4x = 100 - 28 

-8x² = 72

x² = 72 : -8

x² = -9 

=> \(x\in\varphi\)

Chắc z á :v ~ 

a, 3( x + 2) - 6( x - 5 ) = 2( 5 - 2x ) 

3x + 6 - 6x + 30 - 10 + 4x = 0

<=> x( 3 - 6 + 4 ) + 6 + 30 - 10 = 0

=> x + 6 + 30 - 10 = 0

=> x= -26

\(\left(2x\right)\left(-4x\right)+28=100\)

\(\Rightarrow8x^2+28=100\)

\(\Rightarrow8x^2=72\)

\(\Rightarrow x^2=72:8\)

\(\Rightarrow x^2=9\)

\(\Rightarrow x=\sqrt{9}=3\)

a, 3(x+2)-6(x-5)=2(5-2x)

    3x+6-6x+30=10-4x

    3x+6-6x+30-10+4x=0

    3x-6x+4x+6+30-10=0

    x+26=0

    x= -26

b, (-2x)(-4x)+28=100

     8x² + 28=100

     8x² = 72

       x²  = 9

     => x=3 hoặc x= -3

a/ 3(x+2) - 6(x-5) = 2(5-2x)

<-> 3x+6-6x+30 = 10-4x

<-> x = -26

b/  (-2x)(-4x) + 28 = 100

<-> 8x = 72

<-> x =  9

1: A=(x-1)^2>=0

Dấu = xảy ra khi x=1

5: B=-(x^2+6x+10)

=-(x^2+6x+9+1)

=-(x+3)^2-1<=-1

Dấu = xảy ra khi x=-3

2: B=x^2+4x+4-9

=(x+2)^2-9>=-9

Dấu = xảy ra khi x=-2

6: =-(x^2-5x-3)

=-(x^2-5x+25/4-37/4)

=-(x-5/2)^2+37/4<=37/4

Dấu = xảy ra khi x=5/2

3: =x^2+x+1/4-1/4

=(x+1/2)^2-1/4>=-1/4
Dấu = xảy ra khi x=-1/2

7: =4x^2+4x+1-2

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

Dấu = xảy ra khi x=-1/2

1: A=(x-1)^2>=0

Dấu = xảy ra khi x=1

5: B=-(x^2+6x+10)

=-(x^2+6x+9+1)

=-(x+3)^2-1<=-1

Dấu = xảy ra khi x=-3

2: B=x^2+4x+4-9

=(x+2)^2-9>=-9

Dấu = xảy ra khi x=-2

6: =-(x^2-5x-3)

=-(x^2-5x+25/4-37/4)

=-(x-5/2)^2+37/4<=37/4

Dấu = xảy ra khi x=5/2

3: =x^2+x+1/4-1/4

=(x+1/2)^2-1/4>=-1/4
Dấu = xảy ra khi x=-1/2

7: =4x^2+4x+1-2

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

Dấu = xảy ra khi x=-1/2

a) |5/3 - x| - |-5/6| = |-5/9|

=> |5/3 - x| - 5/6 = 5/9

=> |5/3 - x| = 5/9 + 5/6

=> |5/3 - x| = 25/18

=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)

a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)

\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)

b, \(\left|x+\frac{1}{102}\right|+\left|x+\frac{2}{102}\right|+...+\left|x+\frac{100}{102}\right|\ge0\)

\(\Rightarrow102x\ge0\Leftrightarrow x\ge0\)=> Ta có thể phá dấu GTTĐ

\(\Rightarrow x+\frac{1}{102}+x+\frac{2}{102}+...+x+\frac{100}{102}=102x\)

\(\Rightarrow100x+\frac{1+2+3+...+100}{102}=102x\Rightarrow2x\Rightarrow x.\)

4x(x-5)-(x-1)(4x-3) = 5

4x² - 20x - [4x² - 3x - (4x - 3)] = 5

4x² - 20x - [4x² - 3x - 4x + 3] = 5

4x² - 20x - 4x_² + 3x + 4x - 3 = 5

-13x = 5 + 3

-13x = 8

x = \(\dfrac{\left(-8\right)}{13}\)

4x(x - 5) - (x - 1)(4x - 3) = 5

<=> 4x² - 20x - 4x² + 3x + 4x - 3 = 5

<=> 4x² - 4x² - 20x + 3x + 4x = 5 + 3

<=> -13x = 8

<=> \(x=-\dfrac{8}{13}\)