Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{24^2+BC^2-30^2}{2\cdot24\cdot BC}=cos60=\dfrac{1}{2}\)

=>\(BC^2-324=24\cdot BC\)

=>\(BC^2-24\cdot BC-324=0\)

=>\(BC^2-24BC+144-468=0\)

=>\(\left(BC-12\right)^2=468\)

=>\(\left[{}\begin{matrix}BC-12=\sqrt{468}=6\sqrt{13}\\BC-12=-6\sqrt{13}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}BC=6\sqrt{13}+12\left(nhận\right)\\BC=12-6\sqrt{13}=6\sqrt{4}-6\sqrt{13}< 0\left(loại\right)\end{matrix}\right.\)

Vậy: \(BC=6\sqrt{3}+12\simeq33,63\left(cm\right)\)

Xét ΔABC có \(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)

=>\(\dfrac{30}{sin60}=\dfrac{24}{sinC}\)

=>\(sinC=24\cdot\dfrac{sin60}{30}=\dfrac{2\sqrt{3}}{5}\)

=>\(\widehat{C}\simeq44^0\)

\(\left\{{}\begin{matrix}-2x+ay=4\\3x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5-3x\\-2x+a\left(5-3x\right)=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5-3x\\-2x+5a-3xa=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-3x+5\\x\left(-3a-2\right)=4-5a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-3x+5\\x\left(3a+2\right)=5a-4\end{matrix}\right.\left(1\right)\)

TH1: \(a=-\dfrac{2}{3}\)

(1) sẽ tương đương với \(\left\{{}\begin{matrix}y=-3x+5\\x\cdot0=5\cdot\dfrac{-2}{3}-4=-\dfrac{10}{3}-\dfrac{12}{3}=-\dfrac{22}{3}\left(vôlý\right)\end{matrix}\right.\)

=>Loại

TH2: a<>-2/3

(1): \(\left\{{}\begin{matrix}y=-3x+5\\x\left(3a+2\right)=5a-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5a-4}{3a+2}\\y=-3x+5=\dfrac{-3\cdot\left(5a-4\right)}{3a+2}+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5a-4}{3a+2}\\y=\dfrac{-15a+12+15a+10}{3a+2}=\dfrac{22}{3a+2}\end{matrix}\right.\)

x>0 và y>0

=>\(\left\{{}\begin{matrix}\dfrac{5a-4}{3a+2}>0\\\dfrac{22}{3a+2}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3a+2>0\\\dfrac{5a-4}{3a+2}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>-\dfrac{2}{3}\\\dfrac{5a-4}{3a+2}>0\end{matrix}\right.\)

\(\dfrac{5a-4}{3a+2}>0\)

TH1: \(\left\{{}\begin{matrix}5a-4>0\\3a+2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>\dfrac{4}{5}\\a>-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow a>\dfrac{4}{5}\)

mà a>-2/3

nên \(a>\dfrac{4}{5}\)

TH2: \(\left\{{}\begin{matrix}5a-4< 0\\3a+2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a< \dfrac{4}{5}\\a< -\dfrac{2}{3}\end{matrix}\right.\)

=>\(a< -\dfrac{2}{3}\)

mà a>-2/3

nên \(a\in\varnothing\)

Vậy: \(a>\dfrac{4}{5}\) 

mà a là số nguyên nhỏ nhất

nên a=1

`-(-x+13-142) +18=55`

`=> x - 13 +142=55-18`

`=> x - 13 +142=37`

`=> x-13=37-142`

`=>x-13=-105`

`=>x=-105+13`

`=>x=-92`

- (- \(x\) + 13 - 142 ) + 18 = 55

    \(x\) - 13 + 142  + 18  = 55

                               \(x\)  = 55  - 18 - 142  + 13

                               \(x\)  = (55 + 13) - (18 + 142)

                                 \(x\) = 68 - 160

                                  \(x\) = - 92 

a: Xét ΔABM và ΔACM có

AB=AC

BM=CM

AM chung

Do đó: ΔABM=ΔACM

=>\(\widehat{BAM}=\widehat{CAM}\)

Xét ΔPAM vuông tại P và ΔQAM vuông tại Q có

AM chung

\(\widehat{PAM}=\widehat{QAM}\)

Do đó: ΔPAM=ΔQAM

=>PA=QA và MP=MQ

b: AP=AQ

=>A nằm trên đường trung trực của PQ(1)

MP=MQ

=>M nằm trên đường trung trực của PQ(2)

Từ (1) và (2) suy ra AM là đường trung trực của PQ

=>AM\(\perp\)PQ

\(\lim\limits\left(\sqrt[3]{1+2n-n^3}-n\right)\)

\(=\lim\limits\dfrac{1+2n-n^3-n^3}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\cdot\sqrt[3]{1+2n-n^3}+n^2}\)

\(=\lim\limits\dfrac{1+2n-2n^3}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\cdot\sqrt[3]{1+2n-n^3}+n^2}\)

\(=\lim\limits\dfrac{n^3\left(-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}\right)}{n^2\cdot\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+n^2\cdot\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+n^2}\)

\(=\lim\limits\dfrac{n\left(-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}\right)}{\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+1}\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits n=+\infty\\\lim\limits\dfrac{-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}}{\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+1}=\dfrac{-2}{1+1+1}=-\dfrac{2}{3}< 0\end{matrix}\right.\)

Phương pháp: Nhân cả tử và mẫu với biểu thức liên hợp

1. \(A\pm B\) có liên hợp là \(A\mp B\).

2. \(\sqrt{A}\pm B\) có liên hợp là \(\sqrt{A}\mp B\).

3. \(\sqrt{A}\pm\sqrt{B}\) có liên hợp là \(\sqrt{A}\mp\sqrt{B}\).

4. \(\sqrt[3]{A}\pm B\) có liên hợp là \(\sqrt[3]{A^2}\mp B\sqrt[3]{A}+B^2\).

Bài giải: Áp dụng biểu thức liên hợp số 4

    \(\lim\left(\sqrt[3]{1+2n-n^3}-n\right)\)

\(=\lim\dfrac{\left(\sqrt[3]{1+2n-n^3}-n\right)\left[\sqrt[3]{\left(1+2n-n^3\right)^2}+n\sqrt[3]{1+2n-n^3}+n^2\right]}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\sqrt[3]{1+2n-n^3}+n^2}\)

\(=\lim\dfrac{1+2n-n^3-n^3}{\sqrt[3]{n^6-4n^4-2n^3+4n^2+4n+1}+\sqrt[3]{n^3+2n^4-n^6}+n^2}\)

\(=\lim\dfrac{\left(1+2n-2n^3\right)\div n^3}{\left(\sqrt[3]{n^6-4n^4-2n^3+4n^2+4n+1}+\sqrt[3]{n^3+2n^4-n^6}+n^2\right)\div n^3}\)

\(=\lim\dfrac{\dfrac{1}{n^3}+\dfrac{2}{n^2}-2}{\sqrt[3]{\dfrac{1}{n^3}-\dfrac{4}{n^5}-\dfrac{2}{n^6}+\dfrac{4}{n^7}+\dfrac{4}{n^8}+\dfrac{1}{n^9}}+\sqrt[3]{\dfrac{1}{n^6}+\dfrac{2}{n^5}-\dfrac{1}{n^3}}+\dfrac{1}{n}}\)

\(=-\infty\)

(vì \(\lim\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-2\right)=-2\) và \(\lim\left(\sqrt[3]{\dfrac{1}{n^3}-\dfrac{4}{n^5}-\dfrac{2}{n^6}+\dfrac{4}{n^7}+\dfrac{4}{n^8}+\dfrac{1}{n^9}}+\sqrt[3]{\dfrac{1}{n^6}+\dfrac{2}{n^5}-\dfrac{1}{n^3}}+\dfrac{1}{n}\right)=0\), chia được \(\dfrac{-2}{0}\) nên ra \(-\infty\))