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ĐKXĐ: x>=0
\(\dfrac{2\sqrt{x}-6}{x-\sqrt{x}+1}< 0\)
mà \(x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\) thỏa mãn ĐKXĐ
nên \(2\sqrt{x}-6< 0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
3 x 3 - 8 x 6
= 9 - 48
= - 39
2 x 15 + 6 - 7
= 30 + 6 - 7
= 36 - 7
= 29
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-2\\y\ne-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{x+2}-\dfrac{3y}{y+1}=-4\\\dfrac{x}{x+2}+\dfrac{2y}{y+1}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+4-4}{x+2}-\dfrac{3y+3-3}{y+1}=-4\\\dfrac{x+2-2}{x+2}+\dfrac{2y+2-2}{y+1}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{4}{x+2}-3+\dfrac{3}{y+1}=-4\\1-\dfrac{2}{x+2}+2-\dfrac{2}{y+1}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{4}{x+2}+\dfrac{3}{y+1}=-4-2+3=-6+3=-3\\-\dfrac{2}{x+2}-\dfrac{2}{y+1}=\dfrac{1}{3}-3=-\dfrac{8}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-4}{x+2}+\dfrac{3}{y+1}=-3\\\dfrac{-4}{x+2}-\dfrac{4}{y+1}=-\dfrac{16}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-4}{x+2}+\dfrac{3}{y+1}+\dfrac{4}{x+2}+\dfrac{4}{y+1}=-3+\dfrac{16}{3}\\\dfrac{-4}{x+2}-\dfrac{4}{y+1}=-\dfrac{16}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{7}{y+1}=\dfrac{7}{3}\\\dfrac{1}{x+2}+\dfrac{1}{y+1}=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=3\\\dfrac{1}{x+2}=\dfrac{4}{3}-\dfrac{1}{3}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2\\x=-1\end{matrix}\right.\left(nhận\right)\)
50-(20+40)
=50-60=-10
\(30+\left(31+69\right)-210\)
\(=30+100-210\)
\(=30-110=-80\)
ĐKXĐ: x<>-2
\(\dfrac{x-3}{x+2}>=0\)
TH1: \(\left\{{}\begin{matrix}x-3>=0\\x+2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\x>-2\end{matrix}\right.\)
=>x>=3
TH2: \(\left\{{}\begin{matrix}x-3< =0\\x+2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =3\\x< -2\end{matrix}\right.\)
=>x<-2
Bài 7:
\(\dfrac{x-2}{5}=\dfrac{-2}{2y+1}\)
=>\(\left(x-2\right)\left(2y+1\right)=5\cdot\left(-2\right)=-10\)
mà 2y+1 lẻ
nên \(\left(x-2;2y+1\right)\in\left\{\left(10;-1\right);\left(-10;1\right);\left(2;-5\right);\left(-2;5\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(12;-1\right);\left(-8;0\right);\left(4;-3\right);\left(0;2\right)\right\}\)
Bài 6:
\(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{70}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)
=>\(\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{140}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)
=>\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{770}:\dfrac{2}{3}=\dfrac{101}{770}\cdot\dfrac{3}{2}=\dfrac{303}{1540}\)
=>\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}=\dfrac{1}{308}\)
=>x+3=308
=>x=305
Bài 8:
a: \(\left(2x-1\right)^2+4>=4\forall x\)
=>\(B=\dfrac{20}{\left(2x-1\right)^2+4}< =\dfrac{20}{4}=5\forall x\)
Dấu '=' xảy ra khi 2x-1=0
=>\(x=\dfrac{1}{2}\)
b: \(x^2+1>=1\forall x\)
=>\(\left(x^2+1\right)^2>=1^2=1\forall x\)
=>\(\left(x^2+1\right)^2+5>=1+5=6\forall x\)
=>\(C=\dfrac{10}{\left(x^2+1\right)^2+5}< =\dfrac{10}{6}=\dfrac{5}{3}\forall x\)
Dấu '=' xảy ra khi x=0
Sửa đề: \(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a^2+2ab+b^2\right)-2\left(a+b\right)+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)\cdot1+1^2\)
\(=\left(a+b-1\right)^2\)
a: KHi xét nghiệm viêm gan thì có 2 kết quả có thể xảy ra: Dương tính, Âm tính
b: Xác suất thực nghiệm là:
\(\dfrac{26}{230}=\dfrac{13}{115}\)
A = {20; 30; 40; 50; 60; 70}
A = {x ∈ N|12 < x ≤ 70 và x ⋮ 10}
A = { 20, 30, 40, 50, 60, 70 }
A = { x ϵ N; x ⋮ 10 }