Cho x thỏa mãn
\(\sqrt{x^2-6x+36}\)+\(\sqrt{x^2-6x+64}\)=7
TÌm GTBT : A= \(\sqrt{4x^2-24x+256}\)-2.\(\sqrt{x^2-6x+36}\)
Cho số thực x thỏa mãn \(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)
Tính giá trị biểu thức \(A=\sqrt{4x^2-24x+256}-2\sqrt{x^2-6x+36}\)
1, cho a = \(4+\sqrt{5}\),b=\(4-\sqrt{5}\)
Tính A=\(\left(a^{2018}-8a^{2017}+11a^{2016}\right)+\left(b^{2018}-8b^{2017}+11b^{2016}\right)\)
2, cho \(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+48}=18\)
Tính A=\(\sqrt{4x^2-24x+256}-2\sqrt{x^2-6x+36}\)
Cho\(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)
Tính: \(\sqrt{x^2-6x+64}-\sqrt{x^2-6x+36}\)
Đặt \(A=\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=18\)
\(B=\sqrt{x^2-6x+64}-\sqrt{x^2-6x+36}\)
\(\Rightarrow A.B=\left(x^2-6x+64\right)-\left(x^2-6x+36\right)=28\)
mà \(A=18\Rightarrow B=\frac{28}{18}=\frac{14}{9}\)
Cho số thực x thỏa mãn:\(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+64}=7\)
giải pt
a.\(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
b.\(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
c.\(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6x\)
a,ĐK: x≥4
Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}=4\)
\(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)
b, ĐK: x≥2
Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
d) \(\sqrt{x^2-12x+36}-x=3\)
e) \(\sqrt{x^2-4x+5}-1=x\)
f) \(\sqrt{x^2-6x+9}+x=3\)
h) \(\sqrt{18x}+\sqrt{32x}-14=0\)
k) \(\sqrt{6x-3}+2=\sqrt{3}\)
h: \(\sqrt{18x}+\sqrt{32x}-14=0\)
\(\Leftrightarrow7\sqrt{2x}=14\)
hay x=2
- Tìm GTNN :
c. C = \(\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)
d. D = \(\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)
e. E = \(\frac{1}{2}\)\(\sqrt{x^2}\)+ \(\sqrt{x^2-2x+1}\)
c/ \(C=\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}\)
\(=|3-x|+|x+5|\ge|3-x+x+5|=8\)
d/ \(D=\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)
\(=\sqrt{\left(x-3\right)^2}+\sqrt{4\left(x+3\right)^2}\)
\(=|3-x|+|x+3|+|x+3|\ge|3-x+x+3|+0=6\)
e/ \(2E=\sqrt{x^2}+2\sqrt{x^2-2x+1}\)
\(=\sqrt{x^2}+2\sqrt{\left(x-1\right)^2}\)
\(=|x|+|1-x|+|x-1|\ge|x+1-x|+0=1\)
\(\Rightarrow E\ge\frac{1}{2}\)
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
a) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
b) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
c) \(\sqrt{x^2+6x-9}-2\sqrt{x^2-2x+1}+\sqrt{x^2}=0\)
a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4
=>2*căn(x+5)=4
=>căn (x+5)=2
=>x+5=4
=>x=-1
b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
=>2*căn x-1=16
=>x-1=64
=>x=65
c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)
TH1: \(x\ge3\)
\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)
TH2: \(2\le x< 3\)
\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
TH3: \(0\le x< 2\)
\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
TH4: \(x< 0\)
\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)