giải các phương trình vô tỉ sau
1) \(x^2-x=2004\left(\sqrt{1+16032x}+1\right)\)
2) \(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=2\left(x-1\right)+\left(2x^2-4x+1\right)\)
giải các phương trình cô tỉ sau
1) \(\sqrt{x+1}-\sqrt{\frac{x+1}{x}}-1=0\)
2) \(\left(x^2+2\right)^2+4\left(x+1\right)^3+\sqrt{x^2+2x+5}=\left(2x-1\right)^3+2\)
3) \(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=2\left(x-1\right)^4\left(2x^2-4x+1\right)\)
đặt \(\sqrt{2x-x^2}=a\)
phương trình trở thành:
\(\sqrt{1+a}+\sqrt{1-a}=2\left(1-a^2\right)^2\left(1-2a^2\right)\)
đến đây thì khai triển đi
1/ Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{x}=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a-\frac{a}{b}-1=0\\a^2-b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab=a+b\\\left(a+b\right)\left(a-b\right)=1\end{cases}}\)
Tới đây b làm nốt nhé
Giải phương trình:
\(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=2\left(x-1\right)^4\left(2x^2-4x+1\right)\)
Giải các phương trình sau:
a) \(\sqrt{1-x^2}=x-1\)
b) \(\sqrt{x^2+4x+4}=x-2\)
c) \(\sqrt{\left(2x+4\right)\left(x-1\right)}=x+1\)
d) \(\sqrt{2x^2+4x-1}=x-2\)
a: Ta có: \(\sqrt{1-x^2}=x-1\)
\(\Leftrightarrow1-x^2=x-1\)
\(\Leftrightarrow1-x^2-x+1=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)
\(\Leftrightarrow\left|x+2\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(x\ge-2\right)\\x+2=2-x\left(x< -2\right)\end{matrix}\right.\Leftrightarrow2x=0\)
hay x=0(loại)
giải các phương trình vô tỉ sau
1) \(\sqrt{x+8}=\frac{3x^2+7x+8}{3x+1}\)
2) \(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\)
ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !
câu 1 ) thì đúng
câu 2 sai đề
bài 1 chắc bạn sai đề. Mình lười lắm nên link đây nhé https://diendantoanhoc.net/topic/96618-sqrtx8frac3x27x84x2/
giải phương trình vô tỉ sau
1) \(2\left(x-2\right).\left(\sqrt[3]{4x-4}+\sqrt{2x-2}\right)=3x-1\)
2) \(\left(x-2\right).\left(\sqrt{2x+3}+\sqrt{2x+2}\right)=2x-1\)
giải hộ mình nha các bạn mình cảm ơn nhiều lắm
\(2\left(x-2\right)\left(\sqrt[3]{4x-4}+\sqrt{2x-2}\right)=3x-1\)
\(\Leftrightarrow2\left(x-2\right)\left[\left(\sqrt[3]{4x-4}-2\right)+\left(\sqrt{2x-2}-2\right)\right]+8\left(x-2\right)=3x-1\)
\(\Leftrightarrow2\left(x-2\right)\left[\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\sqrt[3]{4x-4}+4}+\frac{2x-6}{\sqrt{2x-2}+2}\right]+\left(5x-15=0\right)\)
\(\left(x-3\right)\left[\frac{8\left(x-2\right)}{...}+\frac{4\left(x-2\right)}{...}+5\right]=0\Leftrightarrow x=3.\)
Giải phương trình vô tỉ sau:
a, \(\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^6}-\sqrt{\left(1-x\right)^3}\right]=1+\sqrt{1-x^2}\)
b, \(\sqrt{x+1}=x^2+4x+5\)
c, \(\sqrt{x+1}=x^{\text{4}}+4x^2+5\)
d, \(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
Giải các phương trình sau:
1/ \(2x^2-8x+\sqrt{x^2-4x+16}=4\)
2/\(3\left(x^2+2\right)=10\sqrt{x^3+1}\)
3/\(\sqrt{3\left(1-x\right)}-\sqrt{3+x}=2\)
Thấy : \(x^2-4x+16=\left(x-2\right)^2+12>0\forall x\)
P/t \(\Leftrightarrow2\left(x^2-4x+16\right)-36+\sqrt{x^2-4x+16}=0\)
Đặt \(t=\sqrt{x^2-4x+16}>0\) ; khi đó :
\(2t^2+t-36=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-\dfrac{9}{2}\left(L\right)\end{matrix}\right.\)
Với t = 4 hay \(\sqrt{x^2-4x+16}=4\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy ...
Câu 1 bạn trên giải rồi mik k giải nx nha
2/ \(3\left(x^2+2\right)=10\sqrt{x^3+1}\)
\(3\left(x^2-x+1\right)+3\left(x+1\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b\ge0\end{matrix}\right.\)
pt⇔ \(3a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3b=b\\a=3b\end{matrix}\right.\)
Đến đây bạn tự giải tiếp nha
3/ \(\sqrt{3-3x}-\sqrt{3+x}=2\)
\(\left(\sqrt{3-3x}-3\right)-\left(\sqrt{3+x}-1\right)=0\)
\(\dfrac{-3\left(x+2\right)}{\sqrt{3-3x}+3}-\dfrac{x+2}{\sqrt{3+x}+1}=0\)
+) \(x=-2\left(TM\right)\)
+) \(x\ne-2\Rightarrow\dfrac{-3}{\sqrt{3-3x}+3}-\dfrac{1}{\sqrt{3+x}+1}=0\)
Vì VT<0 => ptvn
2 ) ĐK : \(x\ge-1\)
P/t \(\Leftrightarrow9\left(x^2+2\right)^2=100\left(x^3+1\right)\)
\(\Leftrightarrow9x^4+36x^2+36=100x^3+100\)
\(\Leftrightarrow9x^4-100x^3+36x^2-64=0\)
\(\Leftrightarrow\left(x^2-10x-8\right)\left(9x^2-10x+8\right)=0\)
\(\Leftrightarrow x^2-10x-8=0\) ( 9x^2 - 10x + 8 > 0 )
\(\Leftrightarrow x=5\pm\sqrt{33}\) ( t/m )
Vậy ...
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 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.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
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:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)