Mấy bài này căng vậy?

a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x =\(\frac{2}{7}\)

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

~Hok tốt~

a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x = 2/7

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

a, \(\frac{3}{4}x-\frac{1}{5}=\frac{7}{4}x+\frac{11}{5}\)

\(\Rightarrow\frac{3}{4}x-\frac{7}{4}x=\frac{11}{5}+\frac{1}{5}\)

\(\Rightarrow-x=\frac{12}{5}\Rightarrow x=-\frac{12}{5}\)

Vậy ... 

b, \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=16\)

\(\Rightarrow x+1=4\)hoặc \(x+1=-4\)

\(\Rightarrow x=3\) hoặc  \(x=-5\)

Vậy ..

3/4x-7/4x=1/5+11/5

(3/4-7/4).x=12/5

-1.x=12/5

x=12/5:-1

x=-12/5

vậy x=-12/5

a,\(\frac{3}{4}x-\frac{1}{5}=\frac{7}{4}x+\frac{11}{5}\)

\(< =>\frac{3x}{4}-\frac{7x}{4}=\frac{11}{5}+\frac{1}{5}=\frac{12}{5}\)

\(< =>-\frac{4x}{4}=\frac{12}{5}< =>-x=\frac{12}{5}\)

\(< =>x=-\frac{12}{5}\)

b, \(x+\frac{1}{2}=\frac{8}{x}+1\left(đk:x\ne0\right)\)

\(< =>x-\frac{8}{x}=1-\frac{1}{2}=\frac{1}{2}\)

\(< =>\frac{x^2-8}{x}=\frac{1}{2}\)

\(< =>2x^2-16=x\)

\(< =>2x^2-x+16=0\)

\(\Delta=\left(-1\right)^2-4.2.16 < 0\)

Vô nghiệm  :V bài này sai đề

1.\(\left(2x-1\right)\left(x-1\right)=0\)

2. \(\left(x+1\right)\left(2x-5\right)=0\)

3. \(\left(2x-1\right)\left(x+4\right)=0\)

4. Vô nghiệm vì VT > 0 \(\forall\)x

bạn lm chi tiết cho mink ik 

Ta có: 7x2+8xy+7y2=10 (*)

=>4x2+8xy+4y2+3x2+3y2=10

=>4(x+y)2+3(x2+y2)=10

=>3(x2+y2)=10-4(x+y)2

Vậy A lớn nhất khi (x+y)2=0=>x=-y

Amax=10/3

Áp dụng bất đẳng thức Cosy cho 2 số dương ta có:

A=x2+y22xy,

=> Amin khi x=y

Thay vào (*) ta được:

7x2+8x2+7x2=10

=>22x2=10

=>x2=10/22 

=> y2=10/22

=>Amin=10/22+10/22=10/11.

Vậy Amin=10/3<=> x=-y

       Amax=10/11<=>x=y.

k cho mình nhé mọi người

câu 2a) xét (x-1)²> hoặc = 0

(x-1)²+(y+1)²> hoặc bằng 0

(x-1)²+(y+1)²+3> hoặc =3

=> GTNN của biểu thức trên là 3

GIÚP minh vs mai mình nộp rui!!!!!!!!!!!!!!!!!!!!@@@@@@@@@@

Ta có :

\(\left(2x^2-3x+1\right)-\left(2x^2-3x+4\right)=0\)

\(\Leftrightarrow2x^2-3x+1-2x^2+3x-4=0\)

\(\Leftrightarrow-3=0\left(ktm\right)\)

\(\Leftrightarrow x\in\varnothing\)

a, Ta có:

f(a + b) = 10(a + b)
f(a) + f(b) = 10a + 10b = 10(a+ b)

=> f(a + b) = f(a) + f(b)

b, f(x) = x² <=> 10x = x² 

                  <=> x = 10 hoặc x = 0
                

\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)

\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)

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

<=>x=2 hoặc x=-2

Vậy S={-2;2}