Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bạn thử tải app này xem có đáp án không nhé <3 https://giaingay.com.vn/downapp.html
a/ \(A=\frac{x}{2}+\frac{1}{2x}+\frac{5x}{2}\ge2\sqrt{\frac{x}{4x}}+\frac{5}{2}.1=\frac{7}{2}\)
\("="\Leftrightarrow x=1\)
b/ \(B=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2\left(x+1\right)}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)
\("="\Leftrightarrow\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=...\)
c/ \(C=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{5\left(2x-1\right)}{6\left(2x-1\right)}}+\frac{1}{6}=\frac{1+2\sqrt{30}}{6}\)
\("="\Leftrightarrow\left(2x-1\right)^2=30\Rightarrow x=...\)
d/ \(D=x+\frac{4}{x}+4\ge2\sqrt{\frac{4x}{x}}+4=8\)
\("="\Leftrightarrow x^2=4\Rightarrow x=...\)
e/ \(E=\left(x+3\right)\left(5-x\right)\le\frac{1}{4}\left(x+3+5-x\right)^2=16\)
\("="\Leftrightarrow x+3=5-x\Rightarrow x=...\)
f/ \(F=\frac{1}{2}\left(2x+6\right)\left(5-2x\right)\le\frac{1}{8}\left(2x+6+5-2x\right)^2=\frac{121}{8}\)
\("="\Leftrightarrow2x+6=5-2x\Leftrightarrow x=...\)
A= [(2√x√x+3)+√x√x+3+3(√xx−9)]:(2√x−2√x−3−11)[(2xx+3)+xx+3+3(xx−9)]:(2x−2x−3−11)với x>= 0 , x #9
xem gi
co ban nho cua toi
may bi loi unikey thong cam nha moi nguoi
hihi
a) \(\sqrt[3]{x}< 2\Leftrightarrow\left(\sqrt[3]{x}\right)^3< 2^3\Leftrightarrow x< 8\)
b) \(\sqrt[3]{2x-1}>-3\Leftrightarrow\left(\sqrt[3]{2x-1}\right)^3>\left(-3\right)^3\Leftrightarrow2x-1>-27\Leftrightarrow2x>-26\Leftrightarrow x>-13\)
c) \(\sqrt[3]{2-3x}\le1\Leftrightarrow\left(\sqrt[3]{2-3x}\right)^3\le1\Leftrightarrow2-3x\le1\Leftrightarrow3x\ge1\Leftrightarrow x\ge\frac{1}{3}\)
d) \(\sqrt[3]{3-4x}\ge5\Leftrightarrow\left(\sqrt[3]{3-4x}\right)^3\ge5^3\Leftrightarrow3-4x\ge125\Leftrightarrow4x\le-122\Leftrightarrow x\le-\frac{61}{2}\)
Bạn ghi lại đề đi bạn
uh