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a) Ta có:
+) a/2=b/3
=>a=2b/3
+) b/5=c/4
=>c=4b/5
Lại có:
a-b+c=49
=> 2b/3 -b + 4b/5 =49
=> 7b/15==49
=> b= 105
Khi đó:
+) a=2b/3=2.105/3=70
+)c=4b/5=4.105/5=84
Vậy a=70; b=105; c=84...
chúc bạn học tốt
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\dfrac{72-x}{7}=\dfrac{x-4}{9}\)
\(\Rightarrow9\left(72-x\right)=7\left(x-4\right)\)
\(\Rightarrow648-9x=2x-28\)
\(\Rightarrow11x-28=648\)
\(\Rightarrow11x=676\Rightarrow x=\dfrac{676}{11}\)
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x+39=259\)
\(\Rightarrow10x=220\Rightarrow x=22\)
\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=\pm10^2\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\Rightarrow x=6\\x+4=-10\Rightarrow x=-14\end{matrix}\right.\)
\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)
\(\Rightarrow x\left(x+3\right)-1\left(x+3\right)=x\left(x+2\right)-2\left(x+2\right)\)
\(\Rightarrow x^2+3x-x-3=x^2+2x-2x-4\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x-3=-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
\(\dfrac{-4}{13}\cdot\dfrac{5}{17}+\dfrac{-12}{13}\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{-4}{17}\cdot\dfrac{5}{17}+\dfrac{-4}{13}\cdot3\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{-5}{17}+\dfrac{-12}{17}+1\right)\)
\(=\dfrac{4}{13}\left(-1+1\right)=\dfrac{4}{13}\cdot0=0\)
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Mà 1-2y lẻ nên:
\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)
b tương tự.
c) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)
d tương tự
3,
\(M=\dfrac{\dfrac{4}{237}-\dfrac{4}{2371}+\dfrac{4}{23711}}{\dfrac{-5}{237}+\dfrac{5}{2371}-\dfrac{5}{23711}}=\dfrac{\left(-4\right)\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}{5\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}=\dfrac{-4}{5}\)
Vậy \(M=\dfrac{-4}{5}\)
2,
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2011}=\dfrac{2011}{a}=\dfrac{a+b+c+2011}{b+c+2011+a}=\dfrac{a+b+c+2011}{a+b+c+2011}=1\)
\(\dfrac{a}{b}=1\Rightarrow a=b\left(1\right)\\ \dfrac{b}{c}=1\Rightarrow b=c\left(2\right)\)
Từ (1) và (2) ta có: \(a=c\)
\(\Rightarrow a+b-c=a+a-a=a\)
1)
b)
\(A=27^{20}+3^{61}+9^{31}\\ =\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\\ =3^{60}+3^{61}+3^{62}\\ =3^{60}\cdot\left(1+3+3^2\right)\\ =3^{60}\cdot\left(1+3+9\right)\\ =3^{60}\cdot13⋮13\)
Vậy \(A⋮13\)
a,
\(\left(-99\right)^{20}=\left(-99\right)^{2\cdot10}=\left[\left(-99\right)^2\right]^{10}=9801^{10}\\ 9999^{100}=\left(9999^{10}\right)^{10}>\left(9999^{10}\right)^1=9999^{10}\)
Vì \(9801^{10}< 9999^{10}< \left(9999^{10}\right)^{10}=9999^{100}\Rightarrow\left(-99\right)^{20}< 9999^{100}\)
Vậy \(\left(-99\right)^{20}< 9999^{100}\)
1/
a) (-99)20 = 9920
Vì 99 < 9999
20 < 100
Nên 9920 < 9999100
Vậy (-99)20 < 9999100
b) \(A=27^{20}+3^{61}+9^{31}\)
\(=\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\)
\(=3^{60}+3^{61}+3^{62}\)
\(=3^{60}\left(1+3+3^2\right)\)
\(=3^{60}.13⋮13\)
Vậy A chia hết cho 13.
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2011}=\dfrac{2011}{a}=\dfrac{a+b+c+2011}{b+c+2011+a}=1\)
\(\Rightarrow\dfrac{a}{b}=1;\dfrac{b}{c}=1\Rightarrow a=b=c\) (*)
Thay (*) vào a + b - c: a + a - a = a
Vậy a + b - c = a.
3. \(M=\dfrac{\dfrac{4}{237}-\dfrac{4}{2371}+\dfrac{4}{23711}}{-\dfrac{5}{237}+\dfrac{5}{2371}-\dfrac{5}{23711}}\)
\(=\dfrac{4\left(\dfrac{1}{237}-\dfrac{1}{2371}+\dfrac{1}{23711}\right)}{-5\left(\dfrac{1}{237}-\dfrac{1}{2371}+\dfrac{1}{23711}\right)}\)
\(=-\dfrac{4}{5}\)
\(\left(\dfrac{3^2.3^4}{6^5.9}\right)^4:\left(\dfrac{2^4}{64}\right)^{12}\)
=\(\left(\dfrac{3^4}{6^5}\right)^4:\left(\dfrac{16}{64}\right)^{12}\)
=\(\left(\dfrac{1}{96}\right)^4:\left(\dfrac{1}{4}\right)^{12}\)
=\(\dfrac{1}{96^4}:\dfrac{1}{4^{12}}\)
=\(\dfrac{4^{12}}{96^4}\)
\(\sqrt{\dfrac{16}{169}}.\dfrac{-3}{2}.\left(\dfrac{3}{2}+\dfrac{-5}{12}\right):\left(-\dfrac{1}{2}\right)\\ =\left|\sqrt{\left(\pm\dfrac{4}{13}\right)^2}\right|.\dfrac{-3}{2}.\dfrac{13}{12}.\left(-2\right)\\ =\left(\dfrac{4}{13}.\dfrac{13}{12}\right).\left(-2.\dfrac{-3}{2}\right)\\ =\dfrac{4}{12}.3=\dfrac{12}{12}=1\)
a) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)
\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
2)
a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72
\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)
\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)
Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)
Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)
b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12
\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)
\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)
Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)
Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)