Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(=\sqrt{\sqrt{2}-1}-\left(\sqrt{2}-1\right)\sqrt{\sqrt{2}+1}\)
\(=\sqrt{\sqrt{2}-1}\left(1-\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\right)\)
\(=2\sqrt{\sqrt{2}-1}\)
b/ \(\Leftrightarrow x^2-12x+36=6561\)
\(\Leftrightarrow x^2-12x-6525=0\)
\(\Leftrightarrow\left(x-87\right)\left(x+75\right)=0\Rightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)
c/ \(\Leftrightarrow4x^2-12x+9=49\)
\(\Leftrightarrow4x^2-12x-40=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Hai câu b; c đều có thể giải bằng cách sử dụng hằng đẳng thức, nhưng cần phá trị tuyệt đối tốn thời gian, tốt nhất là bình phương cho lẹ
\(\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\sqrt{2}+2}\)
\(Đat:A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\Rightarrow A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=2\left(\sqrt{2}+1\right)\Rightarrow A=\sqrt{2\sqrt{2}+2}\left(vì:\sqrt{\sqrt{2}-1};\sqrt{\sqrt{2}+1}>0\right)\) \(\Rightarrow\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\sqrt{2}+2}=\sqrt{2\sqrt{2}+2}-\sqrt{2\sqrt{2}+2}=0\)
\(b,\sqrt{x^2-12x+36}=\sqrt{\left(x-6\right)^2}=\left|x-6\right|=81\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right..Vậy:x\in\left\{87;-75\right\}\)
\(c,\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=7\Leftrightarrow\left|2x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right..Vậy:x\in\left\{-2;5\right\}\)
câu a và câu b bình phương là ra
câu c vì mỗi dấu căn luôn luôn lớn hơn hoặc bằng 0 nên từng cái căn 1 phải bằng 0tuwf đó tính ra đc x = -3
c)\(pt\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x+3\right)^2}=0\)
Đặt căn (x+3) ra ngoài
Biết đâu làm đó , sai thôi đừngg chửi nhé
1, Rút gọn
a) A = \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\) = \(\dfrac{\left(\sqrt{x}\right)^2+\sqrt{xy}}{\left(\sqrt{y}\right)^2+\sqrt{xy}}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}\) = \(\dfrac{\sqrt{x}}{\sqrt{y}}\)
b) B = \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) . \(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\)
= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}.\dfrac{bc^3}{\left(a-b\right)}}\) = \(\sqrt{\left(a-b\right)^2.b^4.c^2}\)
= \(\left|a-b\right|\) . \(\left|b^2\right|\) . \(\left|c\right|\)
= -(a -b) .b2. c
bài 2:
a/ \(\sqrt{x^2-4}-\sqrt{x-2}=0\) đk: x≥2
<=> \(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
<=>\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
vậy pt có 1 nghiệm x = 2
b/ \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)
Ta có: \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=\sqrt{3\left(x^2+4x+4\right)+4}+\sqrt{\left(y^2-4y+4\right)+9}=\sqrt{3\left(x+2\right)^2+4}+\sqrt{\left(y-2\right)^2+9}\ge\sqrt{4}+\sqrt{9}=2+3=5\)
=> Dấu ''='' xảy ra khi x = -2; y = 2
Vậy pt có nghiệm x=-2; y = 2
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) \(H=\frac{1}{9}x^2+3x-1=\left[\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{9}{2}+\frac{81}{4}\right]-\frac{81}{4}-1=\left(\frac{x}{3}+\frac{9}{2}\right)^2-\frac{85}{4}\ge-\frac{85}{4}\)
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
a)\(\sqrt{x+9}=7\)
Đk:\(x\ge-9\).Bình phương 2 vế của pt ta có:
\(\sqrt{\left(x+9\right)^2}=7^2\)\(\Leftrightarrow x+9=49\Leftrightarrow x=40\)
b)\(\sqrt{x^2-12x+36}=81\)
Đk:\(x\ge6\)
\(\Leftrightarrow\sqrt{\left(x-6\right)^2}=81\)
\(\Leftrightarrow x-6=81\Leftrightarrow x=87\)
c)\(\sqrt{x-1}=4\)
Đk:\(x\ge1\).Bình phương 2 vế của pt ta có:
\(\sqrt{\left(x-1\right)^2}=4^2\)
\(\Leftrightarrow x-1=16\Leftrightarrow x=17\)