bn lấy (2718+1564) x 134

Bn Nguyễn Phạm Lan Chi / Bn trả lời rõ ra cho mình đc ko?                  

\(=27.\left(-53\right)+27.\left(-47\right)\\ =27.\left[\left(-53\right)+\left(-47\right)\right]\\ =27.-100\\ =-2700\)

-2700

b: \(BC\cdot\sin B\cdot\sin C\)

\(=BC\cdot\dfrac{AC}{BC}\cdot\dfrac{AB}{BC}\)

\(=\dfrac{BC\cdot AH\cdot BC}{BC^2}=AH\)

Câu 2:

a: \(\frac{1}{\sqrt{11}+2\sqrt3}+\frac{1}{\sqrt{23+\sqrt{528}}}\)

\(=\frac{2\sqrt3-\sqrt{11}}{\left(2\sqrt3-\sqrt{11}\right)\left(2\sqrt3+\sqrt{11}\right)}+\frac{1}{\sqrt{23+2\cdot\sqrt{132}}}\)

\(=2\sqrt3-\sqrt{11}+\frac{1}{\sqrt{\left(2\sqrt3+\sqrt{11}\right)^2}}=2\sqrt3-\sqrt{11}+\frac{1}{2\sqrt3+\sqrt{11}}\)

\(=2\sqrt3-\sqrt{11}+2\sqrt3-\sqrt{11}=4\sqrt3-2\sqrt{11}\)

b: \(\frac{1}{\sqrt5-\sqrt3}-\frac{1}{\sqrt3-\sqrt2}-\frac{2}{\sqrt5+\sqrt2}\)

\(=\frac{\sqrt5+\sqrt3}{5-3}-\frac{\sqrt3+\sqrt2}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}-\frac{2\left(\sqrt5-\sqrt2\right)}{5-2}\)

\(=\frac12\cdot\sqrt5+\frac12\cdot\sqrt3-\sqrt3-\sqrt2-\frac23\left(\sqrt5-\sqrt2\right)\)

\(=\frac12\sqrt5-\frac12\cdot\sqrt3-\sqrt2-\frac23\cdot\sqrt5+\frac23\cdot\sqrt2=\frac{-1}{6}\sqrt5-\frac13\sqrt2-\frac12\sqrt3\)

c:

x>2

=>x-1>1

=>\(\sqrt{x-1}>1\)

=>\(\sqrt{x-1}-1>0\)

\(\frac{1}{\sqrt{x+2\sqrt{x-1}}}+\frac{1}{\sqrt{x-2\sqrt{x-1}}}\)

\(=\frac{1}{\sqrt{x-1+2\cdot\sqrt{x-1}+1}}+\frac{1}{\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}}\)

\(=\frac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}+\frac{1}{\sqrt{\left(\sqrt{x-1}-1\right)^2}}=\frac{1}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x-1}-1}\)

\(=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-1-1}=\frac{2\cdot\sqrt{x-1}}{x-2}\)

Ta có: \(\left|y+3\right|\ge0\forall y\)

\(\left|2x+y\right|\ge0\forall x,y\)

Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)

\(\frac{3}{x}+\frac{y}{2}=\frac{1}{16}\)

=>\(\frac{6+xy}{2x}=\frac{1}{16}\)

=>16(xy+6)=2x

=>8(xy+6)=x

=>8xy-x=-48

=>x(8y-1)=-48

=>(x;8y-1)∈{(1;-48);(-48;1);(-1;48);(48;-1);(2;-24);(-24;2);(-2;24);(24;-2);(3;-16);(-16;3);(-3;16);(16;-3);(4;-12);(-12;4);(-4;12);(12;-4);(6;-8);(-8;6);(-6;8);(8;-6)}

=>(x;8y)∈{(1;-47);(-48;2);(-1;49);(48;0);(2;-23);(-24;3);(-2;25);(24;-1);(3;-15);(-16;4);(-3;17);(16;-2);(4;-11);(-12;5);(-4;13);(12;-3);(6;-7);(-8;7);(-6;9);(8;-5)}

=>(x;y)∈{\(\left(1;-\frac{47}{8}\right);\left(-48;\frac14\right);\left(-1;\frac{49}{8}\right);\left(48;0\right);\left(2;-\frac{23}{8}\right);\left(-24;\frac38\right);\left(-2;\frac{25}{8}\right);\left(24;-\frac18\right);\left(3;-\frac{15}{8}\right);\left(-16;\frac12\right)\) ; \(\left(4;-\frac{11}{8}\right);\left(-12;\frac58\right);\left(-4;\frac{13}{8}\right);\left(12;-\frac38\right);\left(6;-\frac78\right);\left(-8;\frac78\right)\) ; \(\left(-6;\frac98\right);\left(8;-\frac58\right)\) }

Ta có: \(\frac{a+b}{b+c}=\frac79\)

=>\(\frac{a+b}{7}=\frac{b+c}{9}\)

\(\frac{c+a}{a+b}=\frac87\)

=>\(\frac{a+c}{8}=\frac{a+b}{7}\)

=>\(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}\)

Đặt \(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}=k\)

=>a+b=7k; a+c=8k; b+c=9k

=>a+c-a-b=8k-7k=k; c+b=9k; a+b=7k

=>c-b=k và c+b=9k và a+b=7k

=>c=(k+9k)/2=5k; b=9k-5k=4k; a=7k-4k=3k

\(P=\frac{3ab+2bc+ac}{a^2+2b^2+3c^2}\)

\(=\frac{3\cdot3k\cdot4k+2\cdot4k\cdot5k+3k\cdot5k}{\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2}\)

\(=\frac{36k^2+40k^2+15k^2}{9k^2+32k^2+75k^2}=\frac{36+40+15}{9+32+75}=\frac{91}{116}\)

Ta có: \(\frac{a+b}{b+c}=\frac79\)

=>\(\frac{a+b}{7}=\frac{b+c}{9}\)

\(\frac{c+a}{a+b}=\frac87\)

=>\(\frac{a+c}{8}=\frac{a+b}{7}\)

=>\(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}\)

Đặt \(\frac{a+b}{7}=\frac{a+c}{8}=\frac{b+c}{9}=k\)

=>a+b=7k; a+c=8k; b+c=9k

=>a+c-a-b=8k-7k=k; c+b=9k; a+b=7k

=>c-b=k và c+b=9k và a+b=7k

=>c=(k+9k)/2=5k; b=9k-5k=4k; a=7k-4k=3k

\(P=\frac{3ab+2bc+ac}{a^2+2b^2+3c^2}\)

\(=\frac{3\cdot3k\cdot4k+2\cdot4k\cdot5k+3k\cdot5k}{\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2}\)

\(=\frac{36k^2+40k^2+15k^2}{9k^2+32k^2+75k^2}=\frac{36+40+15}{9+32+75}=\frac{91}{116}\)