Giải pt \(\sqrt{x+1}+\sqrt{2x+3}=5\) ???
Giải pt \(\sqrt{x+1}+\sqrt{2x+3}=5\) ???
đk x\(\ge-1\)
pt \(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=25\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=21-3x\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(21-3x\right)^2\)đk \(x\le7\)
\(\Leftrightarrow8x^2+20x+12=9x^2-126x+441\)
\(\Leftrightarrow x^2-146x+429=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=143\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)
ĐK : \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge-1\)
Ta có :
\(\sqrt{x+1}+\sqrt{2x+3}=5\\ \Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=25\\ \Leftrightarrow x+1+2x+3+2\sqrt{\left(x+1\right)\left(2x+3\right)}=25\\ \Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=25-3x-4\\ \Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=21-3x\\ \Leftrightarrow4\left(x+1\right)\left(2x+3\right)=\left(21-3x\right)^2\\ \Leftrightarrow4\left(2x^2+5x+3\right)=441-126x+9x^2\\ \Leftrightarrow8x^2+20x+12=441-126x+9x^2\\ \Leftrightarrow441-126x+9x^2-8x^2-20x-12=0\\ \Leftrightarrow x^2-146x+429=0\\ \Leftrightarrow\left[{}\begin{matrix}x=143\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\)
Vậy phương trình đã cho có 2 nghiệm là x=143 và x=3
1.\(\sqrt{\frac{\left(1-x\right)}{x}}=\frac{\left(2x+x^2\right)}{1+x^2}\)
2. 3(2-\(\sqrt{x+2}\))=2x+\(\sqrt{x+6}\)
3. \(\sqrt[3]{x+2}+\sqrt[3]{x+1}=\sqrt[2]{2x^2}+\sqrt[3]{2x^2+1}\)
4. \(\sqrt[3]{x+24}+\sqrt{12-x}=6\)
Toán 10 ạ, giúp em với
4. đặt \(\sqrt[3]{x+24}=a\) và \(\sqrt{12-x}=b\)(b>=0)
==>ta có hệ pt
\(\int_{a^3+b^2=36}^{a+b=6}\)<=> \(\int_{a^3+\left(6-a\right)^2=36}^{b=6-a}\)<=> \(\int_{b=6-a}^{a^3+a^2-12a=0}\)<=> \(\int_{b=6-a}^{a\left(a^2+a-12\right)=0}\)<=>\(\int_{b=6-a}^{a\left(a+4\right)\left(a-3\right)=0}\)
đến đây bạn tự tìm a;b rufit hay vào tìm x là ok
3. \(\Leftrightarrow\sqrt[3]{2x^2}-\sqrt[3]{x+1}+\sqrt[3]{2x^2+1}-\sqrt[3]{x+2}=0\)
\(\Leftrightarrow\frac{2x^2-x-1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{2x^2-x-1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}=0\)
\(\Leftrightarrow2x^2-x-1=0\)
( do \(\frac{1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}>0\forall xTMĐK\))
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2=\frac{9}{8}\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{3}{4}\\x-\frac{1}{4}=-\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\) ( TM )
\(\sqrt{x+1}+\sqrt{5-x}=\sqrt{m+4x-x^2}\)
Tìm m để pt có nghiệm .
Lời giải:
ĐKXĐ: \(-1\leq x\leq 5; m\geq x^2-4x\)
\(\Rightarrow m\geq (x^2-4x)_{\max}, \forall x\in [-1; 5]\) hay \(m\geq5\) (1)
Ta có: \(\sqrt{x+1}+\sqrt{5-x}=\sqrt{m+4x-x^2}\)
\(\Rightarrow 6+2\sqrt{(x+1)(5-x)}=m+4x-x^2\) (bình phương hai vế)
\(\Leftrightarrow (-x^2+4x+5)-2\sqrt{(x+1)(5-x)}+1-12+m=0\)
\(\Leftrightarrow m=12-(\sqrt{-x^2+4x+5}-1)^2=f(x)\)
Để pt có nghiệm thì \(f(x)_{\min}\leq m\leq f(x)_{\max}\)
Ta có: \((\sqrt{-x^2+4x+5}-1)^2\geq 0, \forall x\in [-1;5]\Rightarrow f(x)\leq 12\) hay \(f(x)_{\max}=12\)
Mặt khác: \(0\leq -x^2+4x+5=9-(x-2)^2\leq 9\)
\(\Rightarrow 0\leq \sqrt{-x^2+4x+5}\leq 3\)
\(\Rightarrow (\sqrt{-x^2+4x+5}-1)^2\leq 4\)
\(\Rightarrow f(x)\geq 12-4=8\Leftrightarrow f(x)_{\min}=8\)
Suy ra để pt có nghiệm thì \(8\leq m\leq 12(2)\)
Do đó từ (1) và (2) suy ra \(8\leq m\leq 12\).
X^6 - 7X^2 +√6 =0
(\(\dfrac{1}{\sqrt{2}}\)+1)/(1-\(\dfrac{1}{\sqrt{2}}\))
\(=\dfrac{\sqrt{2}+1}{\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\left(\sqrt{2}+1\right)^2\)
\(=3+2\sqrt{2}\)
giải bất phương trình sau
1)\(\sqrt{x+3}>3-\sqrt{x+4}\)
2) \(\sqrt{x+2}-\sqrt{x}+1< \sqrt{x}\)
giải phương trình
\(3x+2+\sqrt{4-x^2}-3\sqrt{2-x}+6\sqrt{2+x}=0\)
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3x+3y\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
Xét VT của (1):
\(3VT\)
\(=\sqrt{5x^2+2xy+2y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{2x^2+2xy+5y^2}.\sqrt{2^2+2^2+1^2}\)
\(=\sqrt{\left(x+y\right)^2+4x^2+y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{\left(x+y\right)^2+x^2+4y^2}.\sqrt{2^2+2^2+1^2}\)
\(\ge\left[2\left(x+y\right)+4x+y\right]+\left[2\left(x+y\right)+x+4y\right]=9x+9y\)
\(\Rightarrow VT\ge3x+3y=VT\)
Đẳng thức xảy ra \(\Leftrightarrow...\Leftrightarrow x=y\)
Sau đó thay \(y=x\) vào pt (2) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left(2x^2-\sqrt{3x+1}\right)+\left(x-5-2\sqrt[3]{19x+8}\right)=0\)
\(\Leftrightarrow\dfrac{4x^2-3x-1}{2x^2+\sqrt{3x+1}}+\dfrac{\left(x+5\right)^3-8\left(19x+8\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4x+1\right)}{2x^2+\sqrt{3x+1}}+\dfrac{ \left(x-1\right)\left(x^2+16x-61\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\dfrac{4x+1}{2x^2+\sqrt{3x+1}}+\dfrac{x^2+16x-61}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}\right]=0\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
\/x^2-x-2 -2×\/x-2 +2=\/x+1
3(√2x+1+√x−2x+11)=4√2x2+x
Dat \(\sqrt{2x+1}=a,\sqrt{x}=b=>x=b^2\)
=> a.b=\(\sqrt{2x^2+x}\)
pt<=> 3(a+2b2+11)=4ab
=> 3a+6b2+33=4ab