Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
)
\(\sqrt{1-x-2x^2}=\sqrt{\left(1+x\right)\left(1-2x\right)}\le\dfrac{1+x-2x+1}{2}=\dfrac{-x+2}{2}\)
(AM-GM)
do đó \(A\le\dfrac{x}{2}+\dfrac{-x+2}{2}=1\)
Dấu = xảy ra khi 1+x=1-2x <=> x=0 (tmđk)
ĐKXĐ: x>0
\(\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=1\)
a) \(\sqrt{x-3}\) xác định
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy..
b) \(\sqrt{3-2x}\) xác định
\(\Leftrightarrow3-2x\ge0\)
\(\Leftrightarrow x\le-\dfrac{3}{2}\)
Vậy..
c) \(\sqrt{4x^2-1}\) xác định
\(\Leftrightarrow4x^2-1\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Rightarrow x\ge\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\le\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow x\le\dfrac{-1}{2}\)
Vậy ...
d) \(\sqrt{3x^2+2}\) xác định
\(\Leftrightarrow3x^2+2\ge0\)
mà \(3x^2\ge0\)
\(\Rightarrow3x^2+2>0\)
Vậy...
e) \(\sqrt{2x^2+4x+5}\) xác định
\(\Leftrightarrow2x^2+4x+5\ge0\)
mà \(2x^2+4x\ge0\)
\(2x\left(x+2\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\le0\\x+2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le-2\end{matrix}\right.\)\(\Rightarrow x\le-2\)
\(\Rightarrow2x^2+4x+5>0\)
Vậy...
( Câu này không chắc lắm nha )
Bài 2: Tách sẵn ra cho bạn luôn nhé, không thì bạn nhấn máy tính ra cũng được :v
a) \(-\dfrac{7}{9}\sqrt{\left(-27\right)^2+6\sqrt{1}}\)
\(=-\dfrac{7}{9}\sqrt{\left(-3\right)^2.\left(-9\right)^2+6}\)
\(=\dfrac{-7}{9}\sqrt{735}\)
\(=\dfrac{-7}{9}\sqrt{49.15}\)
\(=\dfrac{-49\sqrt{15}}{9}\)
b) \(\sqrt{49}\sqrt{12^2}+\sqrt{256}:\sqrt{8^2}\)
\(=84+2=86\)
c)\(\sqrt{\left(\sqrt{3-1}\right)^2-\sqrt{\left(\sqrt{3+1}\right)^2}}\)
\(=\sqrt{2-2}\)
= 0
\(\dfrac{\sqrt{\dfrac{-\left(2\right)^5}{5^3.5^2}.\dfrac{-\left(5\right)^3}{2^9}.5^2}}{\sqrt[3]{\dfrac{-\left(3\right)^3}{2^6}.\dfrac{\left(5\right)^2}{3^2.2^5}.\dfrac{\left(5\right)^4}{3^4}}}=\dfrac{\sqrt{\dfrac{1}{2^4}}}{\sqrt[3]{\dfrac{-\left(5\right)^6}{2^{12}.3^3}}}=\dfrac{\dfrac{1}{4}}{\sqrt[3]{\left(\dfrac{-5^2}{2^4.3}\right)^3}}=\dfrac{\dfrac{1}{4}}{\dfrac{-25}{48}}=\dfrac{-12}{25}\)
a) \(\sqrt{\left|x\right|-1}\) biểu thức sau có nghĩa \(\Leftrightarrow\) \(\left|x\right|-1\ge0\)
\(\Leftrightarrow\left|x\right|\ge1\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\hoac\\x\le-1\end{matrix}\right.\)
b) \(\sqrt{\left|x-1\right|-3}\) biểu thức sau có nghĩa \(\Leftrightarrow\left|x-1\right|-3\ge0\)
\(\Leftrightarrow\left|x-1\right|\ge3\) \(\left\{{}\begin{matrix}x-1\ge3\\hoac\\x-1\le-3\end{matrix}\right.\)
c) \(\sqrt{4-\left|x\right|}\) biểu thức sau có nghĩa \(\Leftrightarrow4-\left|x\right|\ge0\)
\(\Leftrightarrow4\ge\left|x\right|\) \(\Leftrightarrow-4\le x\le4\)
a) Đặt \(t=\sqrt{2x^2-3x+5}\ge0\) thì
\(2t=t^2-11\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1+2\sqrt{3}\\t=1-2\sqrt{3}\end{matrix}\right.\)
Vì \(t\ge0\) nên \(t=1+2\sqrt{3}\)
\(\Rightarrow\sqrt{2x^2-3x+5}=1+2\sqrt{3}\)
\(\Leftrightarrow2x^2-3x+5=13-4\sqrt{3}\)
\(\Leftrightarrow2x^2-3x-8+4\sqrt{3}=0\)
Giải pt trên tìm được x
c) ĐK: \(x\ge0\)
Đặt \(a=\sqrt{x}\ge0;b=\sqrt{x+3}\ge0\)
pt trên đc viết lại thành
\(2b^2+2ab=4\left(a+b\right)\)
\(\Leftrightarrow\left(b-2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{x}=-\sqrt{x+3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=x+3\end{matrix}\right.\)
Vậy pt có 1 nghiệm duy nhất x = 1.
b) ĐK: tự làm
Ta có \(\left(x+5\right)\left(2-x\right)=-x\left(x+3\right)+10\)
Đặt \(a=\sqrt{x}\ge0;b=\sqrt{x+3}\ge0\)
pt trên đc viết lại thành
\(-a^2b^2+10=3ab\)
\(\Leftrightarrow-a^2b^2-3ab+10=0\) (*)
Đặt \(t=ab\ge0\) thì (*) \(\Rightarrow-t^2-3t+10=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ab=t=2\\ab=t=-5\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}=2\)
Bạn tự làm tiếp nhé
Dùng BĐT Bunhiacopski:
Ta có: \(ac+bd\le\sqrt{a^2+b^2}.\sqrt{c^2+d^2}\)
Mà \(\left(a+c\right)^2+\left(b+d\right)^2\)
\(=a^2+b^2+2\left(ac+bd\right)+c^2+d^2\)
\(\le\left(a^2+b^2\right)+2\sqrt{a^2+b^2}.\sqrt{c^2+d^2}+c^2+d^2\)
\(\Rightarrow\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\le\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\) (Đpcm)
Câu hỏi của Hoàng Khánh Linh - Toán lớp 8 - Học toán với OnlineMath copy nhớ ghi nguồn
giải giùm mình với, mình sẽ tặng 1SP
1. \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
2. a) Với a>b>0 thì
\(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\dfrac{a-\sqrt{a^2-b^2}}{b}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b^2}{b\sqrt{a^2-b^2}}=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b}{\sqrt{a^2-b^2}}\)
\(=\dfrac{a-b}{\sqrt{a^2-b^2}}=\dfrac{a-b}{\sqrt{a-b}.\sqrt{a+b}}=\sqrt{\dfrac{a-b}{a+b}}\)
b) Thay a = 3b ta được
\(Q=\sqrt{\dfrac{a-b}{a+b}}=\sqrt{\dfrac{3b-b}{3b+b}}=\sqrt{\dfrac{2b}{4b}}=\sqrt{\dfrac{1}{2}}=\dfrac{\sqrt{2}}{2}\)
1) d) ta có : \(VT=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)
\(\Rightarrow\) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\) (đpcm)
.Cảm ơn!!
đk : \(x\ne4\) ; \(x\ge0\)
1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\) = \(\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{3}{2+\sqrt{x}}\)
b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)
\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)
c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4
ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên
\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3
mà \(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3
\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)
vậy x = 1 nguyên thì Q nguyên
2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)
= \(4\sqrt{a}-5\sqrt{10a}\)
b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)
c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-2\sqrt{22}-11+3\sqrt{22}\)
= \(22+\sqrt{22}\)