tìm giá trị nhỏ nhất:
\(a,\sqrt{\left(x-2000\right)^2}+\sqrt{\left(x-2001\right)^2}\)
\(b,\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+4\right)^2}+\sqrt{\left(x+5\right)^2}\)
\(c,\sqrt{\left(2x-1\right)^2}+\sqrt{\left(3x+4\right)^2}\)
tìm giá trị nhỏ nhất của
A=\(\sqrt{\left(x+2\right)^2}+\sqrt{\left(x+3\right)^2}=5\)
B=\(\sqrt[]{x+2\sqrt{x-1}+\sqrt{x-2\sqrt{x-1}}}\)
C=\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x+\sqrt{4x-1}}\)
1.
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$
Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$
$\Leftrightarrow (x+2)(x+3)\leq 0$
$\Leftrightarrow -3\leq x\leq -2$
2. ĐKXĐ: $x\geq 1$
\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|\geq |\sqrt{x-1}+1+1-\sqrt{x-1}|=2\)
Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$
$\Leftrightarrow 1-\sqrt{x-1}\geq 0$
$\Leftrightarrow 0\leq x\leq 2$
3.
$C\sqrt{2}=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}$
$=2\sqrt{(4x-1)+2\sqrt{4x-1}+1}=2\sqrt{(\sqrt{4x-1}+1)^2}$
$=2|\sqrt{4x-1}+1|$
Vì $\sqrt{4x-1}\geq 0$ nên $|\sqrt{4x-1}+1|\geq 1$
$\Rightarrow C\sqrt{2}\geq 2$
$\Rightarrow C\geq \sqrt{2}$
Vậy $C_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=\frac{1}{4}$
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
a. \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
b. \(3\left(2+\sqrt{x}\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
c. \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
d. \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)
\(=x-3\sqrt{x}-5\sqrt{x}-15\)
\(=x-8\sqrt{x}-15\)
b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)
\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)
\(=-x+2\sqrt{x}+12\)
c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)
\(=x-9-5\sqrt{x}+5\)
\(=x-5\sqrt{x}-4\)
d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3\sqrt{x}-6-x+1\)
\(=-x+3\sqrt{x}-5\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Timg GTNN của các bt:
a. \(A=\left|x-2000\right|+\left|x-2001\right|+\left|x-2002\right|\)
b. \(\sqrt{\left(x-8\right)^2}+\sqrt{\left(x-9\right)^2}+\sqrt{\left(x-10\right)^2}+\sqrt{\left(x+11\right)^2}\)
a) \(\left|x-2000\right|+\left|x-2002\right|=\left|x-2000\right|+\left|2002-x\right|\)
\(\ge\left|x-2000+2002-x\right|=2\) (1)
Dấu "=" \(\Leftrightarrow\left(x-2000\right)\left(2002-x\right)\ge0\)
\(\Leftrightarrow2000\le x\le2002\)
+ \(\left|x-2001\right|\ge0\forall x\). "=" \(\Leftrightarrow x=2001\) (2)
Từ (1) và (2) suy ra \(A\ge2\)
Dấu "=" \(\Leftrightarrow x=2001\)
b) \(B=\left|x-8\right|+\left|x-9\right|+\left|x-10\right|+\left|x+11\right|\)
+ \(\left|x-10\right|+\left|x+11\right|=\left|x+11\right|+\left|10-x\right|\)
\(\ge\left|x+11+10-x\right|=21\) (3)
Dấu "=" \(\Leftrightarrow\left(x+11\right)\left(10-x\right)\ge0\Leftrightarrow-11\le x\le10\)
+ \(\left|x-8\right|+\left|x-9\right|\ge\left|x-8+9-x\right|=1\) (4)
"=" \(\Leftrightarrow\left(x-8\right)\left(9-x\right)\ge0\Leftrightarrow8\le x\le9\)
Từ (3) và (4) suy ra \(B\ge22\)
"=" \(\Leftrightarrow8\le x\le9\)