Bài 5: Gọi số người của đội A,đội B,đội C lần lượt là a(người),b(người),c(người)

(Điều kiện: \(a,b,c\in Z^+\))

Tổng số người của ba đội là 130 người nên a+b+c=130

Số cây mỗi người của đội A;B;C trồng được lần lượt là 2;3;4 cây nên 2a=3b=4c

=>\(\dfrac{2a}{12}=\dfrac{3b}{12}=\dfrac{4c}{12}\)

=>\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}\)

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

\(\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b+c}{6+4+3}=\dfrac{130}{13}=10\)

=>\(a=10\cdot6=60;b=4\cdot10=40;c=3\cdot10=30\)

Vậy: số người của đội A,đội B,đội C lần lượt là 60(người),40(người),30(người)

Bài 8:

a: \(\dfrac{x}{5}=\dfrac{y}{6}\)

=>\(\dfrac{x}{20}=\dfrac{y}{24}\)

\(\dfrac{y}{8}=\dfrac{z}{7}\)

=>\(\dfrac{y}{24}=\dfrac{z}{21}\)

Do đó: \(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=k\)

=>x=20k;y=24k;z=21k

x+y-z=69

=>20k+24k-21k=69

=>23k=69

=>k=3

=>\(x=20\cdot3=60;y=24\cdot3=72;z=21\cdot3=63\)

b: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

=>x=3k;y=4k;z=5k

\(2x^2+2y^2-3z^2=-100\)

=>\(2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

=>\(k^2=4\)

=>\(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1: k=2

=>\(x=3\cdot2=6;y=4\cdot2=8;z=5\cdot2=10\)

TH2: k=-2

=>\(x=3\cdot\left(-2\right)=-6;y=4\cdot\left(-2\right)=-8;z=5\cdot\left(-2\right)=-10\)

Bài 10:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)

a:

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)

\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)

b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)

 \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)

\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)

\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)

Bài 10:

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)

a:

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)

\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)

b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)

 \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)

\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)

\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)

Bài 14:

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

\(A=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\cdot\left(-x\right)\cdot\left(-y\right)\)

=-xyz

=-2

26. 

\(a.\left(\dfrac{3}{7}\right)^5\cdot x=\left(\dfrac{3}{7}\right)^7\\ =>x=\left(\dfrac{3}{7}\right)^7:\left(\dfrac{3}{7}\right)^5\\ =>x=\left(\dfrac{3}{7}\right)^{7-5}\\ =>x=\left(\dfrac{3}{7}\right)^2\\ =>x=\dfrac{9}{49}\\ b.\left(0,09\right)^3x=-\left(0,09\right)^2\\ =>\left[\left(0,3\right)^2\right]^3\cdot x=-\left[\left(0,3\right)^2\right]^2\\ =>x=-\left(0,3\right)^4:\left(0,3\right)^6\\ =>x=-\left(0,3\right)^{-2}\\ =>x=-\left(\dfrac{10}{3}\right)^2\\ =>x=-\dfrac{100}{9}\)

27:

a: Vì \(0< \dfrac{1}{2}< 1\)

và 40<50

nên \(\left(\dfrac{1}{2}\right)^{40}>\left(\dfrac{1}{2}\right)^{50}\)

b: \(243^3=\left(3^5\right)^3=3^{15};125^5=\left(5^3\right)^5=5^{15}\)

mà 3<5

nên \(243^3< 125^5\)

\(\dfrac{5^4.18^4}{125.9^5.16}\\ =\dfrac{5^4.\left(2.3^2\right)^4}{5^3.\left(3^2\right)^5.2^4}\\ =\dfrac{5^4.2^4.3^8}{5^3.2^4.3^{10}}\\ =\dfrac{5}{3^2}\\ =\dfrac{5}{9}\)

\(\dfrac{11}{5}-\left(0,35+x\right)=1\dfrac{1}{2}\\ \dfrac{11}{5}-\left(\dfrac{7}{20}+x\right)=\dfrac{3}{2}\\ \dfrac{11}{5}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{44}{20}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{37}{20}-x=\dfrac{3}{2}\\ x=\dfrac{37}{20}-\dfrac{3}{2}\\ x=\dfrac{7}{20}\)

\(\dfrac{15}{34}+\dfrac{15}{17}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)

\(=\dfrac{15}{34}+\dfrac{19}{34}+\dfrac{15}{17}-1-\dfrac{15}{17}+\dfrac{2}{3}\)

\(=1-1+\dfrac{2}{3}=\dfrac{2}{3}\)

a: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)

mà \(\widehat{bMB}=50^0\)

nên \(\widehat{NMC}=50^0\)

Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)

=>\(\widehat{MNC}+110^0=180^0\)

=>\(\widehat{MNC}=70^0\)

Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)

=>\(\widehat{C}+50^0+70^0=180^0\)

=>\(\widehat{C}=60^0\)

b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)

=>\(\widehat{NMB}+50^0=180^0\)

=>\(\widehat{NMB}=130^0\)

Ta có: MN//AB

=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)

=>\(\widehat{CBA}=50^0\)

BN là phân giác của góc CBA

=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)

=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)

c: BN là phân giác của góc CBA

=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAN}+60^0+50^0=180^0\)

=>\(\widehat{BAN}=70^0\)

Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)

=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)