Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{1}{3};x_2=3\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
cho đáp án tự làm (vì cách lm của mik bị ném đá khá nhiều lần òi :D)
\(x=-1\)
c) như câu b nhé :D
\(x=-2004\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
Bài 1:
\(a,\dfrac{x}{3}=\dfrac{y}{7}\) và \(x+y=20\)
\(=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\Rightarrow x=2.3=6\)
\(y=2.7=14\)
Vậy \(x=6\) và \(y=14\)
\(b,\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=6\)
\(=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\Rightarrow x=2.5=10\)
\(y=2.2=4\)
Vậy \(x=10\) và \(y=4\)
\(c,\dfrac{x}{7}=\dfrac{18}{14}\)
Từ tỉ lệ thức trên ta có:
\(14x=7.18\)
\(x=\dfrac{7.18}{14}\)
\(x=9\)
Vậy \(x=9\)
\(d,6:x=1\dfrac{3}{4}:5\)
\(6:x=\dfrac{7}{20}\)
\(x=6:\dfrac{7}{20}\)
\(x=\dfrac{120}{7}\)
Vậy \(x=\dfrac{120}{7}\)
\(e,\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(x-y+z=8\)
\(=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)
\(\Rightarrow x=2.2=4\)
\(y=2.4=8\)
\(z=2.6=12\)
Vậy \(x=4;y=8;z=12\)
a, \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{1}{2}\)
Từ đó suy ra x=1,5; y=3,5
b,\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{1}{2}\)
Từ đó suy ra x=2,5; y=1
c,\(\dfrac{x}{7}=\dfrac{18}{14}\Leftrightarrow\dfrac{x}{7}=\dfrac{9}{7}\Rightarrow x=9\)
d,\(\dfrac{6}{x}=\dfrac{\dfrac{7}{4}}{5}\Leftrightarrow\dfrac{6}{x}=\dfrac{24}{7}\left(\dfrac{\dfrac{7}{4}}{5}\right)\Leftrightarrow\dfrac{6}{x}=\dfrac{6}{\dfrac{120}{7}}\Rightarrow x=\dfrac{120}{7}\)
e,\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{8}=\dfrac{x-y+z}{2-4+8}=\dfrac{4}{3}\)
Từ đó suy ra x=\(\dfrac{8}{3}\); y=\(\dfrac{16}{3}\); z=\(\dfrac{32}{3}\)
a. Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{5}=\dfrac{y}{7}=\dfrac{y-2x}{7-5}=\dfrac{24}{2}=12\)
\(\Rightarrow2x=12\cdot5=60\Rightarrow x=60:2=30\)
\(y=12\cdot7=84\)
Vậy x = 30 ; y = 84
b. Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+3y}{3+2\cdot3}=\dfrac{18}{9}=2\)
\(\Rightarrow x=2\cdot3=6\)
\(y=2\cdot2=4\)
Vậy x = 6 ; y = 4
c. Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
\(\Rightarrow x=2\cdot2=4\)
\(y=3\cdot2=6\)
\(z=4\cdot2=8\)
Vậy x = 4 ; y = 6 ; z = 8
d. Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-y-z}{2-3-4}=\dfrac{15}{-5}=-3\)
\(\Rightarrow x=-3\cdot2=-6\)
\(y=-3\cdot3=-9\)
\(z=-3\cdot4=-12\)
Vậy \(x=-4;y=-6;z=-8\)
Bài 1.
Giải
a) Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-12+21}{n-4}=\dfrac{3\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để \(A\in Z\) thì \(\dfrac{21}{n-4}\in Z\)
\(\Rightarrow21⋮\left(n-4\right)\)
\(\Rightarrow\left(n-4\right)\inƯ\left(21\right)\)
\(\Rightarrow\left(n-4\right)\in\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Ta có bẳng sau:
Vậy \(n\in\left\{-17;-3;1;3;5;7;11;25\right\}\) thì \(A\in Z.\)
b) Ta có: \(B=\dfrac{6n+5}{2n-1}=\dfrac{6n-3+8}{2n-1}=\dfrac{3\left(2n-1\right)+8}{2n-1}=3+\dfrac{8}{2n-1}\)
Để \(B\in Z\) thì \(\dfrac{8}{2n-1}\in Z\)
\(\Rightarrow8⋮\left(2n-1\right)\)
\(\Rightarrow\left(2n-1\right)\inƯ\left(8\right)\)
\(\Rightarrow\left(2n-1\right)\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Ta có bảng sau:
Vậy \(n\in\left\{\dfrac{-7}{2};\dfrac{-3}{2};\dfrac{-1}{2};0;1;\dfrac{3}{2};\dfrac{5}{2};\dfrac{9}{2}\right\}\)
Bạn Nguyen Thi Huyen giải bài 1 rồi nên mình giải tiếp các bài kia nhé!
Bài 2:
\(\dfrac{x-18}{2000}+\dfrac{x-17}{2001}=\dfrac{x-16}{2002}+\dfrac{x-15}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-18}{2000}-1\right)+\left(\dfrac{x-17}{2001}-1\right)=\left(\dfrac{x-16}{2002}-1\right)+\left(\dfrac{x-15}{2003}-1\right)\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}=\dfrac{x-2018}{2002}+\dfrac{x-2018}{2003}\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}-\dfrac{x-2018}{2002}-\dfrac{x-2018}{2003}=0\)
\(\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Dễ thấy \(\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}\) nên:
\(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\ne0\). Do đó:
\(x-2018=0\Leftrightarrow x=2018\)
Bài 3:
a) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{20}{4x}+\dfrac{xy}{4x}=\dfrac{20+xy}{4x+4x}=\dfrac{20+xy}{8x}=\dfrac{1}{8}\)
Hoán vị ngoại tỉ ta có: \(\dfrac{20+xy}{8x}=\dfrac{1}{8}\Leftrightarrow\dfrac{8}{8x}=\dfrac{1}{x}=\dfrac{1}{8}\Leftrightarrow x=8\)
Thế x = 8 vào : \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\) .Ta có: \(\dfrac{5}{8}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{y}{4}=\dfrac{1}{8}-\dfrac{5}{8}=\dfrac{-2}{4}\). Ta có: \(\dfrac{y}{4}=\dfrac{-2}{4}\Leftrightarrow y=-2\)
Vậy: \(\left[{}\begin{matrix}x=8\\y=-2\end{matrix}\right.\)
b) \(\dfrac{1}{x}-\dfrac{2}{y}=\dfrac{3}{1}\Rightarrow\dfrac{y}{x}-2=\dfrac{3}{1}\) (hoán vị ngoại tỉ)
\(\Leftrightarrow\dfrac{y}{x}=\dfrac{5}{1}\). Suy ra nghiệm x,y có dạng \(\left[{}\begin{matrix}x=1k\\y=5k\end{matrix}\right.\left(k\in Z\right)\). Bằng các phép thử lại ta dễ dàng suy ra x,y vô nghiệm.