Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(x\ge\frac{1}{6}\)
2.\(x\le0\)
3.\(4-5x\ge0\Leftrightarrow x\le\frac{4}{5}\)
4.mọi x
Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ
Mn giúp em vs ạ! Thanks trước!
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
a) \(A=\sqrt{81}.\sqrt{\frac{9}{4}}+2\sqrt{16}-3=\sqrt{9^2}.\sqrt{\left(\frac{3}{2}\right)^2}+2\sqrt{4^2}-3=9.\frac{3}{2}+2.4-3=\frac{37}{2}\)
b) \(B=\sqrt{9-2\sqrt{14}}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)
c) Không rút gọn được.
Bài 2 : Mình hướng dẫn thôi nhé ^^
a) \(M=x^2-10x+30=\left(x^2-10x+25\right)+5=\left(x-5\right)^2+5\ge5\)
b) \(N=4x^2-12x+1=\left[\left(2x\right)^2-12x+9\right]-8=\left(2x-3\right)^2-8\ge-8\)
c) \(P=x^2-x-1=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
d) \(Q=16x^2-8x+3=\left[\left(4x\right)^2-8x+1\right]+2=\left(4x-1\right)^2+2\ge2\)
e) \(H=\frac{1}{9}x^2+3x-1=\left[\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{9}{2}+\frac{81}{4}\right]-\frac{81}{4}-1=\left(\frac{x}{3}+\frac{9}{2}\right)^2-\frac{85}{4}\ge-\frac{85}{4}\)
Trả lời:
a, \(\sqrt{x^2-8x+16}=4\)
\(\Rightarrow x^2-8x+16=4^2\)
\(\Leftrightarrow x^2-8x=0\)
\(\Leftrightarrow x\left(x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)
Vậy x = 0; x = 8 là nghiệm của pt.