Có: \(3a=2b\Rightarrow\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15}\)

      \(4b=5c\Rightarrow\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\)

=> \(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{-a-b+c}{-10-15+12}=\frac{-52}{-13}=4\)

=>\(\frac{a}{10}=4\Rightarrow a=40\)

     \(\frac{b}{15}=4\Rightarrow b=60\)

     \(\frac{c}{12}=4\Rightarrow c=48\)

Bài 1:

\(A=x^2y-y+xy^2-x=\left(x^2y+xy^2\right)-\left(x+y\right)\\ =xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

Voqis x=-1;y=3 ta có:

\(A=\left(-1+3\right)\left(-1\cdot3-1\right)=2\cdot\left(-4\right)=-8\)

b) \(B=x^2y^2+xy+x^3+y^3=\left(x^2y^2+x^3\right)+\left(xy+y^3\right)\\ =x^2\left(y^2+x\right)+y\left(x+y^2\right)=\left(x+y^2\right)\left(x^2+y\right)\)

Với x=-1;y=3 ta có:

\(B=\left(-1+3^2\right)\left(-1^2+3\right)=8\cdot2=16\)

c) \(C=2x+xy^2-x^2y-2y=\left(2x-2y\right)+\left(xy^2-x^2y\right)\\ =2\left(x-y\right)+xy\left(y-x\right)=\left(x-y\right)\left(2-xy\right)\)

Với x=-1;y=3 ta có:

\(C=\left(-1-3\right)\left(2-\left(-1\right)\cdot3\right)=-4\cdot5=-20\)

d) phân tích tt

\(x^2+3x-18=0\)

\(\Leftrightarrow x^2-3x+6x-18=0\)

\(\Leftrightarrow x\left(x-3\right)+6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-6\end{array}\right.\)

Vậy x=3;x=-6

thêm đề đi

Có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}=\frac{y}{2}=\frac{x+y}{3+2}=\frac{20}{5}=4\)

=> \(\frac{x}{3}=4\Rightarrow x=12\)

    \(\frac{y}{2}=4\Rightarrow y=8\)

a) \(A=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2=\left(2x-1+3x+1\right)^2=\left(5x\right)^2=25x^2\)

Thay \(x=-\frac{1}{5}\) vào A ta cọ

\(A=25\cdot\left(-\frac{1}{5}\right)^2=25\cdot\frac{1}{25}=1\)

b)\(B=\left(4x-3\right)^2-2\left(16x^2-9\right)+\left(4x+3\right)^2\)

\(=\left(4x-3\right)^2-2\left(4x-3\right)\left(4x+3\right)+\left(4x+3\right)^2=\left(4x-3-4x-3\right)^2=-6^2\)

Vói x=-17 thì \(B=\left(-6\right)^2=36\)

\(f\left(4\right)=3\cdot4+1=13\)

\(\sqrt{\left(-5\right)^2}=\sqrt{25}=5\)

\(\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=5+9-5\cdot0,7=10,5\)