Bài 1: Tìm x
a, \(\sqrt{4x+1}\) = 4
b, \(\sqrt{3-x}\) = 2
c, \(\sqrt{x+1}\) + \(\frac{1}{2}\sqrt{4x+4}\) = 6 - \(\frac{1}{3}\sqrt{9x+9}\)
d, \(\sqrt{4x^2-12x+9}\) - x = 5
e, \(\sqrt{16x^2+8x+1}\) = 10
Bài 2: Tìm x,y,z
x + y + z = 2\(\sqrt{x}\)+ 2\(\sqrt{y-3}\) + 2\(\sqrt{z}\)
Bài 3: Cho x < y < 0
Rút gọn \(\sqrt{x^2}+\sqrt{y^2}-\sqrt{x^2-2xy+y^2}\)
Bài 4: Tìm GTNN
a, x - 2\(\sqrt{x}\) + 3
b, \(\sqrt{x-4\sqrt{y}+13}\)
c, \(\sqrt{2x-4\sqrt{y}+6}\)
d, \(-\frac{4}{x^2+2x+5}\)
Bài 5: Cho A = \(\frac{3\sqrt{x}+11}{\sqrt{x}+2}\)
Tìm x ϵ Z để A nguyên
4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
ĐÁP ÁN
Bài 1:
a. x= \(\frac{15}{4}\)
b. x= -1
c. x= 34
d. x= 8
e. x= \(\frac{9}{4}\)
Bài 3: Rút gọn
\(\sqrt{x^2}+\sqrt{y^2}-\sqrt{x^2-2xy+y^2}\)= -2x
BIẾT ĐC CÓ 2 BÀI À!!!
1.a)\(\sqrt{4x+1}=4\)
\(\Leftrightarrow4x+1=16\)
\(\Leftrightarrow4x=15\)
\(\Leftrightarrow x=\frac{15}{4}\)
b)\(\sqrt{3-x}=2\)
\(\Leftrightarrow3-x=4\)
\(\Leftrightarrow x=-1\)
c)\(\sqrt{x+1}+\frac{1}{2}\sqrt{4x+4}=6-\frac{1}{3}\sqrt{9x+9}\)
\(\Leftrightarrow\sqrt{x+1}+\frac{1}{2}\sqrt{4\left(x+1\right)}+\frac{1}{3}\sqrt{9\left(x+1\right)}=6\)
\(\Leftrightarrow\sqrt{x+1}+\frac{1}{2}.2\sqrt{x+1}+\frac{1}{3}.3\sqrt{x+1}=6\)
\(\Leftrightarrow\sqrt{x+1}+\sqrt{x+1}+\sqrt{x+1}=6\)
\(\Leftrightarrow3\sqrt{x+1}=6\)
\(\Leftrightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
d)\(\sqrt{4x^2-12x+9}-x=5\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}-x=5\)
\(\Leftrightarrow\left|2x-3\right|-x=5\)
-TH: \(2x-3\ge0\Leftrightarrow x\ge\frac{3}{2}\) thì ta được phương trình:
\(2x-3-x=5\)
\(\Leftrightarrow x=8\)(chọn)
-TH: \(2x-3< 0\Leftrightarrow x< \frac{3}{2}\) thì ta được phương trình:
\(3-2x-x=5\)
\(\Leftrightarrow-3x=2\)
\(\Leftrightarrow x=-\frac{2}{3}\)(chọn)
e)\(\sqrt{16x^2+8x+1}=10\)
\(\Leftrightarrow\sqrt{\left(4x+1\right)^2}=10\)
\(\Leftrightarrow\left|4x+1\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=10\\4x+1=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{4}\\x=-\frac{11}{4}\end{matrix}\right.\)
3.\(\sqrt{x^2}+\sqrt{y^2}-\sqrt{x^2-2xy+y^2}\)
\(=\left|x\right|+\left|y\right|-\left|x-y\right|\)
\(=-x-y-\left(y-x\right)\)
\(=-x-y+x-y\)
\(=-2y\)
5.Ta có:
\(A=\frac{3\sqrt{x}+11}{\sqrt{x}+2}=\frac{3\sqrt{x}+6+5}{\sqrt{x}+2}=3+\frac{5}{\sqrt{x}+2}\)
Để A nguyên thì \(\frac{5}{\sqrt{x}+2}\) cũng nguyên
\(\Rightarrow5⋮\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}+2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{-7;-3;-1;3\right\}\)
\(\Rightarrow x\in\left\{9\right\}\)
