a) Đặt \(A=|x-2|-12\)

Ta có: \(|x-2|\ge0\forall x\)

\(\Rightarrow|x-2|-12\ge0-12\)

Hay \(A\ge-12\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\)

                        \(\Leftrightarrow x=2\)

Vậy Min A =-12 \(\Leftrightarrow x=2\)

xem lại cho anh phần b 

\(25\%.y+50\%.y-\frac{3}{4}.y+4.y=10\)

\(y.\left(\frac{1}{4}+\frac{1}{2}-\frac{3}{4}+4\right)=10\)

\(y.4=10\)

\(y=\frac{5}{2}\)

\(x.\frac{1}{4}-\frac{3}{4}=6:\frac{3}{4}\)

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

\(x.\frac{1}{4}=8+\frac{3}{4}\)

\(x.\frac{1}{4}=\frac{35}{4}\)

\(x=\frac{35}{4}:\frac{1}{4}\)

\(x=35\)

25% x y + 50% x y - 3/4 x y + 4 x y = 10

    1/4 x y + 1/2 x y - 3/4 x y + 4 x y = 10

                 y x ( 1/4 + 1/2 - 3/4 + 4 ) = 10

                                                y x 4 = 10

                                                y       = 10 : 4

                                                y       = 2.5

a ) =0

b) = -1

c) =-1

d) =4

e) =11

oi bi chi

Tìm \(x\) câu a:

\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0

\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0

\(\frac{11}{15}x\) = - \(\frac25\)

\(x=-\frac25:\frac{11}{15}\)

\(x\) = - \(\frac25\times\frac{15}{11}\)

\(x\) = - \(\frac{6}{11}\)

Vậy \(x=-\frac{6}{11}\)

Tìm \(x\) câu b:

\(x\) x 25% = 0,5

\(x\times0,25\) = 0,5

\(x=0,5:0,25\)

\(x=2\)

Vậy \(x=2\)

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )