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\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)
\(2\sqrt{x}-1>0\);\(4x>0\)
\(\Rightarrow x>0\)thì \(A>A^2\)
a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)
= \(\sqrt{\left(2a^2-3\right)^2}-\sqrt{\left(a^2-4\right)^2}\)
= \(|2a^2-3|-|a^2-4|\)
= \(2a^2-3+a^2-4\)
= \(3a^2-7\)
Thay a=\(\sqrt{3}\).Ta có:
\(3.\left(\sqrt{3}\right)^2-7\)
= 3.3-7=2
b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)
= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)
= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)
= \(|a\sqrt{10}-6|\)
= \(-a\sqrt{10}+6\)
Thay a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:
\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)
= -3+6 =3
b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}\)
\(=4x-\sqrt{8}+\frac{x\left(x+2\right)}{x+2}\)
\(=4x-\sqrt{8}+x\)
\(=5x-\sqrt{8}\)
Với \(x=-\sqrt{2}\) ta có:
\(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)
1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
a/ \(A=\sqrt{\left(a-4\right)^2}-3a=\left|a-4\right|-3a\)
+) với a<4: A = 4-a-3a=4-4a
+)với a≥4: A = a-4-3a=-2a - 4
Với a = -3 <4 => A = 4 - 4 . (-3) = 16
b/ \(B=\sqrt{\left(1-2x\right)^2}-2x=\left|1-2x\right|-2x\)
+) nếu x \(\le\frac{1}{2}\) :
\(B=1-2x-2x=-4x+1\)
+) nếu \(x>\frac{1}{2}:B=2x-1-2x=-1\)
với \(x=-\frac{3}{2}< \frac{1}{2}\Rightarrow B=-4\cdot\left(-\frac{3}{2}\right)+1=7\)
c/đk: \(x\ne\pm4\)
\(C=\frac{\sqrt{\left(2x-1\right)^2}}{\left(x-4\right)\left(x+4\right)}\cdot\left(x-4\right)^2=\frac{\left|2x-1\right|\cdot\left(x-4\right)}{x+4}\)
+) nếu \(x\ge\frac{1}{2}:B=\frac{\left(2x-1\right)\left(x-4\right)}{x+4}\)
+) nếu \(x< \frac{1}{2}:B=\frac{-\left(2x-1\right)\left(x-4\right)}{x+4}\)
Với \(x=7\left(>\frac{1}{2}\right):B=\frac{\left(2\cdot7-1\right)\cdot\left(7-4\right)}{7+4}=\frac{39}{11}\)