a) \(y+30\%y=-1.3\)

\(y+\dfrac{3}{10}y=\dfrac{-13}{10}\)

\(y\left(1+\dfrac{3}{10}\right)=\dfrac{-13}{10}\)

\(y\left(\dfrac{10}{10}+\dfrac{3}{10}\right)=\dfrac{-13}{10}\)

\(y\cdot\dfrac{13}{10}=\dfrac{-13}{10}\)

\(y=\dfrac{-13}{10}:\dfrac{13}{10}\)

\(y=\dfrac{-13}{10}\cdot\dfrac{10}{13}\)

\(y=-1\)

Vậy \(y=-1\).

b) \(y-25\%y=\dfrac{1}{2}\)

\(y-\dfrac{1}{4}y=\dfrac{1}{2}\)

\(y\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\)

\(y\left(\dfrac{4}{4}-\dfrac{1}{4}\right)=\dfrac{1}{2}\)

\(y\cdot\dfrac{3}{4}=\dfrac{1}{2}\)

\(y=\dfrac{1}{2}:\dfrac{3}{4}\)

\(y=\dfrac{1}{2}\cdot\dfrac{4}{3}\)

\(y=\dfrac{2}{3}\)

Vậy \(y=\dfrac{2}{3}\).

c) \(3\dfrac{1}{3}y+16\dfrac{3}{4}=-13.25\)

\(\dfrac{10}{3}y+\dfrac{67}{4}=\dfrac{-53}{4}\)

\(\dfrac{10}{3}y=\dfrac{-53}{4}-\dfrac{67}{4}\)

\(\dfrac{10}{3}y=-30\)

\(y=-30:\dfrac{10}{3}\)

\(y=-30\cdot\dfrac{3}{10}\)

\(y=-9\)

Vậy \(y=-9\).

a)

Vậy .

b)

Vậy .

c)

Vậy .

bài 1:vì 2010:2=1005 nên ta chia A làm ngoặc 2 số liên tiếp

<=> A= \(\left(2^1+2^2\right)+\left(2^3+2^4\right)+.\ldots+\left(2^{2019}+2^{2010}\right)\)

\(A=2\left(2+1\right)+2^3\left(2+1\right)+\cdots+2^{2019}\left(2+1\right)\)

\(A=2\cdot3+2^3\cdot3+\cdots+2^{2019}\cdot3\)

\(A=3\left(2^1+2^3+\cdots+2^{2019}\right)\)

=> A⋮3

bài 2:

a) sửa đề: so sánh A với B= \(2^{2011}-1\)

<=> \(2A=2^1+2^2+.\ldots+2^{2011}\)

\(\Rightarrow2A-A=\left(2^1+2^2+.\ldots+2^{2011}\right)-\left(2^0+2^1+\cdots+2^{2010}\right)\)

\(A=2^{2011}-1\)

vậy A=B

b) vì 2019 và 2011 đều> 2010 nên A>B

c) ƯCLN(440,300)=20

=> \(3^{440}=\left(3^{22}\right)^{20}\)

\(5^{300}=\left(5^{15}\right)^{20}\)

mà ta có: \(3^{22}=3^2\cdot3^{20}=9\cdot\left(3^4\right)^5=9\cdot81^5\)

\(5^{15}=\left(5^3\right)^5=125^5\)

vì \(9\cdot81^5>125^5\)

=> A>B

bài 3:

d) xy+3x-7y=21

xy+3x-7y-21=0

x(y+3)-7(y+3)=0

(y+3)(x-7)=0

=> x=7 và y=-3

bài 4:

=> \(A^2=ab-bc-ca+bc\)

\(A^2=ab-ca\)

\(A^2=a\left(b-c\right)=-20\cdot-5=100\)

TH1: A=10

TH2: A=-10

d) \(x.\left(y+2\right)-y=15\)

\(\Rightarrow x.\left(y+2\right)=15+y\)

\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)

y + 2 là ước nguyên của 13

\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)

\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)

\(y+2=13\Rightarrow y=11\Rightarrow x=2\)

\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)

Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu

a)\(y+30\%y=-1,3\Rightarrow y+\frac{3}{10}y=-1,3\Rightarrow y\left(1+\frac{3}{10}\right)=-1,3\Rightarrow y\times1,3\)\(=-1,3\Rightarrow y=-1\)

b)\(y-25\%y=\frac{1}{2}\Rightarrow y-\frac{1}{4}y=\frac{1}{2}\Rightarrow y\left(1-\frac{1}{4}\right)=\frac{1}{2}\Rightarrow y\times\frac{3}{4}=\frac{1}{2}\Rightarrow y=\frac{1}{2}:\frac{3}{4}\Rightarrow y=\frac{2}{3}\)

c)\(3\frac{1}{3}y+16\frac{3}{4}=13,25\Rightarrow\frac{10}{3}y+\frac{67}{4}=\frac{53}{4}\Rightarrow\frac{10}{3}y=\frac{53}{4}-\frac{67}{4}\Rightarrow\frac{10}{3}y=\frac{-7}{2}\Rightarrow y=\frac{-21}{20}\)

-21/20

a) ta có :     y + 30/100y = -1,3

           hay       y(1+ 3/10)    = - 13/10

                       y.13/10 = -13/10

                       y= -13/10:13/10

                       y= -1

b)       y - 25/100y=1/2

hay    y(1-1/4)=1/2

         y.3/4 = 1/2

         y=2/3

c)    10/3y + 16,75 = -13,25

       10/3y= -13,25 - 16,75

       10/3y= -30

             y= - 30:10/3

             y= -9

a)

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)

Áp dụng tc của dãy tỉ só bằng nhau

\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)

=> x=2.10=20

    y=5.10=50

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)

     \(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)

Mà 2;5 cùng dấu

=> x; y cùng dấu

Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)

 Mình không biết.