a) 

Ta có : vì|1/2-1/3+x| lớn hơn hoặc bằng 0 

Còn -1/4-|y| bé hơn hoặc bằng 0

=> ko tồn tại x

b) 

Ta có: |x-y| lớn hơn hoặc bằng 0 và|y+9/25| lớn hơn hoặc bằng 0 mà:

| x-y|+ |y+9/25| =0 => |x-y| =0 và |y+9/25|=0

 Xét |y+9/25| có:

| y+9/25|=0 => y+9/25=0 => y=-9/25

Thay y = -9/25 vào |x-y| =0 => x=-9/25

 Vậy x=y=-9/25

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

\(a.\frac{Y}{3}=\frac{27}{Y}\Leftrightarrow\frac{Y^2}{3Y}=\frac{3.27}{3Y}\)

\(\Rightarrow Y^2=3.27=81\)

Do \(\orbr{\begin{cases}81=\left(-9\right)^2\\81=9^2\end{cases}\Rightarrow}\orbr{\begin{cases}Y=-9\\Y=9\end{cases}}\)

Vậy \(Y\in\left\{-9;9\right\}.\)

\(b.\frac{15}{Y}=\frac{Y}{9}\Leftrightarrow\frac{15.9}{Y.9}=\frac{Y^2}{9.Y}\)

\(\Rightarrow15.9=135=Y^2\)

Do \(\orbr{\begin{cases}135=\left(-\sqrt{135}\right)^2\\135=\sqrt{135}^2\end{cases}\Rightarrow}\orbr{\begin{cases}Y=-\sqrt{135}\\Y=\sqrt{135}\end{cases}}\)

Vậy \(Y\in\left\{-\sqrt{135};\sqrt{135}\right\}.\)

\(c.\frac{15}{48}:y=\frac{10}{4}\)

\(\Rightarrow y=\frac{15}{48}:\frac{10}{4}=\frac{1}{8}\)

Vậy \(y=\frac{1}{8}.\)

\(d.\frac{20}{y}.9=\frac{60}{135}\)

\(\Rightarrow\frac{20}{y}=\frac{60}{135}:9=\frac{20}{135}\)

\(\Leftrightarrow y=135\)

Vậy \(y=135.\)

\(e.\frac{16}{Y}=\frac{Y}{9}\Leftrightarrow\frac{16.9}{Y.9}=\frac{Y^2}{Y.9}\)

\(\Rightarrow16.9=144=Y^2\)

Do \(\orbr{\begin{cases}144=\left(-12\right)^2\\144=12^2\end{cases}\Rightarrow\orbr{\begin{cases}Y=-12\\Y=12\end{cases}}}\)

Vậy \(Y\in\left\{-12;12\right\}\)

\(g.\frac{25}{Y}=\frac{Y}{9}\Leftrightarrow\frac{25.9}{Y.9}=\frac{Y^2}{9.Y}\)

\(\Rightarrow25.9=225=Y^2\)

Do \(\orbr{\begin{cases}225=\left(-15\right)^2\\225=15^2\end{cases}\Rightarrow\orbr{\begin{cases}Y=-15\\Y=15\end{cases}}}\)

Vậy \(Y\in\left\{-15;15\right\}.\)

\(h.y-\frac{12}{42}=\frac{10}{12}\)

\(\Rightarrow y=\frac{10}{12}+\frac{12}{42}=\frac{47}{42}\)

Vậy \(Y=\frac{47}{42}.\)

\(i.y+\frac{5}{30}=\frac{18}{2}\)

\(\Rightarrow y=\frac{18}{2}-\frac{5}{30}=\frac{53}{54}\)

Vậy \(y=\frac{53}{54}.\)

mấy câu này gần như là giống nhau nên mik giải 1 câu rồi bạn dựa theo mà giải tiếp nha

y/3 = 27/y => y² = 3 . 27

               => y² = 81 => y = 9 hoặc -9

vậy .....

\(a,5\frac{1}{4}+3,25-50\%+y=15,25\)

\(\Leftrightarrow5,25+3,25-\frac{50}{100}+y=15,25\)

\(\Leftrightarrow5,25+3,25-0,5+y=15,25\)

\(\Leftrightarrow8+y=15,25\)

\(\Leftrightarrow y=15,25-8=7,25\)

\(b,(y-3):2=2010\)

\(\Leftrightarrow y-3=4020\)

\(\Leftrightarrow y=4023\)

\(c,\frac{7+y}{33}=\frac{9}{11}\)

\(\Leftrightarrow7+y=\frac{9}{11}\cdot33\)

\(\Leftrightarrow7+y=\frac{9}{1}\cdot3\)

\(\Leftrightarrow7+y=27\Leftrightarrow y=20\)

a: A=y(x-4)-5(x-4)

=(x-4)(y-5)

Khi x=14 và y=5,5 thì A=(14-4)(5,5-5)=0,5*10=5

b: \(B=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Khi x=5,2 và y=4,8 thì B=(5,2+4,8)(5,2-5)

=0,2*10=2

d: Khi x=5,75 và y=4,25 thì

D=5,75^3-5,75^2*4,25+4,25^3

=8087/64

bài 1

A)37/60

B)13/45

C)1/4

bài 2

a)42/25:y/5=5/6

y=10,08

b)27/y:9/4=3/7

y=28

a: \(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|-2.5\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.5+3.75=1.25=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{5}{4}\\x+\dfrac{4}{15}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{59}{60}\\x=-\dfrac{91}{60}\end{matrix}\right.\)

c: \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

d: Ta có: \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{5}{4}\right)=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{3}{2}\right\}\)