Tìm x:
\(\sqrt{2x^2-4x+3}+\sqrt{\left(x-1\right)^2.\left(x+5\right)^2}\) = 2x - \(x^2\) -2
Những câu hỏi liên quan
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, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
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 $(x+2)(-x-3)\geq 0$
$\Leftrightarrow (x+2)(x+3)\leq 0$
$\Leftrightarrow -3\leq x\leq -2$
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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 $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$
$\Leftrightarrow 1-\sqrt{x-1}\geq 0$
$\Leftrightarrow 0\leq x\leq 2$
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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 $x=\frac{1}{4}$
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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, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải pt :
a, \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^2+2\)
b, \(\sqrt{4x^2+x+6}=4x-2+7\sqrt{x+1}\)
c, \(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)
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, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
giải pt :
a, \(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
b, \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^2+2\)
c, \(\sqrt{4x^2+x+6}=4x-2+7\sqrt{x+1}\)
d, \(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)
giải pt , sqrt{x^4+4x^2}+sqrt{x+x^2}sqrt{left(x^2+sqrt{x}right)^2+9x^2}.x0x^30x^32.0.sqrt{0}x^32xsqrt{x}x^32xsqrt{x} 4left(x^3-2xsqrt{x}right)^204left(x^6-4x^4sqrt{x}+4x^2xright)04x^6-16x^4sqrt{x}+16x^2x04x^6+16x^316x^4sqrt{x}16x^4+4x^5+4x^6+16x^316x^4+4x^5+16x^4sqrt{x}4x^3left(x+1right)left(x^2+4right)4left(4x^4+4x^4sqrt{x}+x^4.xright)4x^3left(x+1right)left(x^2+4right)4left(2x^2+x^2sqrt{x}right)^22sqrt{2x^3left(x+1right)left(x^2+4right)}2left(2x^2+x^2sqrt{x}right)x^4+x^2+4x^2+x+2sqrt{2x^3le...
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giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)
\(x=0\)
\(x^3=0\)
\(x^3=2.0.\sqrt{0}\)
\(x^3=2x\sqrt{x}\)
\(x^3=2x\sqrt{x}\)
\(4\left(x^3-2x\sqrt{x}\right)^2=0\)
\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)
\(4x^6-16x^4\sqrt{x}+16x^2x=0\)
\(4x^6+16x^3=16x^4\sqrt{x}\)
\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)
\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)
\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)
\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)
\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)
vậy x=0 là nghiệm của pt =))
cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?
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Giải phương trình:
1: \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^2+2\)
2: \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{16x-4x^2-15}\)
14 tháng 1 2021 lúc 12:59
Giải phương trình 1, \(x^2+9x+7=\left(2x+1\right)\sqrt{2x^2+4x+5}\)
2, GPT \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
3. GHPT \(\left\{{}\begin{matrix}x^2-2y-1=2\sqrt{5y+8}+\sqrt{7x-1}\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
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2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
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3.
ĐKXĐ: ...
Từ pt dưới:
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow y=x-2\)
Thế vào pt trên:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)
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