Xét ΔAHB vuông tại H có \(tanBAH=\dfrac{BH}{AH}\)

=>\(BH=AH\cdot tanBAH=4\cdot tan28\simeq2,13\left(cm\right)\)

Xét ΔAHC vuông tại H có

\(tanC=\dfrac{AH}{HC}\)

=>\(HC=\dfrac{AH}{tanC}=\dfrac{4}{tan40}\simeq4,77\left(cm\right)\)

ΔAHB vuông tại H

=>\(AH^2+HB^2=AB^2\)

=>\(AB=\sqrt{AH^2+HB^2}\simeq4,53\left(cm\right)\)

ΔAHC vuông tại H

=>\(AH^2+HC^2+AC^2\)

=>\(AC=\sqrt{AH^2+HC^2}\simeq6,23\left(cm\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)

\(P=\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để P=3/2 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{3}{2}\)

=>\(3\left(\sqrt{x}-2\right)=2\sqrt{x}\)

=>\(3\sqrt{x}-2\sqrt{x}=6\)

=>\(\sqrt{x}=6\)

=>x=36(nhận)

a: 

b: Phương trình hoành độ giao điểm là:

-2x-4=x-1

=>-2x-x=-1+4

=>-3x=3

=>x=-1

Thay x=-1 vào y=x-1, ta được:

y=-1-1=-2

Vậy: Tọa độ giao điểm là A(-1;-2)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3y\right)\left(2x-y\right)=0\\6x^2+7xy-5y^2=0\end{matrix}\right.\)

TH1: \(2x-3y=0\Rightarrow y=\dfrac{2}{3}x\) thay vào pt dưới

\(\Rightarrow6x^2+7x.\left(\dfrac{2}{3}x\right)-5\left(\dfrac{2}{3}x\right)^2=0\)

\(\Leftrightarrow\dfrac{76}{9}x^2=0\Rightarrow x=0\Rightarrow y=0\)

TH2: \(2x-y=0\Rightarrow y=2x\)

Tương tự ta cũng được \(x=0;y=0\)

Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(0;0\right)\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}13x^2-39xy+13y^2=-13\\2x^2+xy+3y^2=13\end{matrix}\right.\)

Cộng vế với vế

\(\Rightarrow15x^2-38xy+16y^2=0\)

\(\Leftrightarrow\left(x-2y\right)\left(15x-8y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2y\\x=\dfrac{8}{15}y\end{matrix}\right.\)

Thay vào pt đầu:

- Với \(x=2y\Rightarrow4y^2-6y^2+y^2=-1\)

\(\Rightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{matrix}\right.\)

- Với \(x=\dfrac{8}{15}y\)

\(\Rightarrow\left(\dfrac{8}{15}y\right)^2-3\left(\dfrac{8}{15}y\right).y+y^2=-1\)

\(\Leftrightarrow-\dfrac{71}{225}y^2=-1\Rightarrow y^2=\dfrac{225}{71}\)

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{15}{\sqrt{71}}\Rightarrow x=\dfrac{8}{\sqrt{71}}\\y=-\dfrac{15}{\sqrt{71}}\Rightarrow x=-\dfrac{8}{\sqrt{71}}\end{matrix}\right.\)

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)

=>\(\dfrac{6}{BC}=sin30=\dfrac{1}{2}\)

=>\(BC=6\cdot2=12\left(cm\right)\)

\(\left|A+B\right|< =\left|A\right|+\left|B\right|\)

=>\(\left(\left|A+B\right|\right)^2< =\left(\left|A\right|+\left|B\right|\right)^2\)

=>\(A^2+B^2+2AB< =A^2+B^2+2\left|AB\right|\)

=>2AB<=2|AB|

=>AB<=|AB|(luôn đúng)

Dấu '=' xảy ra khi AB>=0

xét ΔABC vuông tại A có \(cosB=\dfrac{AB}{BC}\)

=>\(\dfrac{6}{BC}=cos30=\dfrac{\sqrt{3}}{2}\)

=>\(BC=6\cdot\dfrac{2}{\sqrt{3}}=4\sqrt{3}\left(cm\right)\)

a: \(\dfrac{3x+5}{2}-x>=1+\dfrac{x+2}{3}\)

=>\(\dfrac{3x+5-2x}{2}>=\dfrac{3+x+2}{3}\)

=>\(\dfrac{x+5}{2}-\dfrac{x+5}{3}>=0\)

=>\(\dfrac{3\left(x+5\right)-2\left(x+5\right)}{6}>=0\)

=>\(\dfrac{x+5}{6}>=0\)

=>x+5>=0

=>x>=-5

b: \(\dfrac{x-2}{3}-x-2< =\dfrac{x-17}{2}\)

=>\(\dfrac{2\left(x-2\right)}{6}+\dfrac{6\left(-x-2\right)}{6}< =\dfrac{3\left(x-17\right)}{6}\)

=>\(2\left(x-2\right)+6\left(-x-2\right)< =3\left(x-17\right)\)

=>\(2x-4-6x-12< =3x-51\)

=>-4x-16<=3x-51

=>-7x<=-35

=>x>=5

c: \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}< =\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)

=>\(\dfrac{4\left(2x+1\right)-3\left(x-4\right)}{12}< =\dfrac{2\left(3x+1\right)-x+4}{12}\)

=>4(2x+1)-3(x-4)<=2(3x+1)-x+4

=>8x+4-3x+12<=6x+2-x+4

=>5x+16<=5x+6

=>16<=6(sai)

Vậy: BPT vô nghiệm

a: \(\dfrac{3\left(2x+1\right)}{20}+1>\dfrac{3x+52}{10}\)

=>\(\dfrac{6x+3}{20}+\dfrac{20}{20}>\dfrac{6x+104}{20}\)

=>6x+23>6x+104

=>23>104(sai)

vậy: \(x\in\varnothing\)

b: \(\dfrac{4x-1}{2}+\dfrac{6x-19}{6}< =\dfrac{9x-11}{3}\)

=>\(\dfrac{3\left(4x-1\right)+6x-19}{6}< =\dfrac{2\left(9x-11\right)}{6}\)

=>12x-3+6x-19<=18x-22

=>-22<=-22(luôn đúng)

Vậy: \(x\in R\)