\(y=\left(x+\sqrt{x^2+1}\right)^3\) chung minh \(\sqrt{x^2+1}.y^'-3y=0\)
Những câu hỏi liên quan
Giải hệ phương trình:
1. left{{}begin{matrix}x+32sqrt{left(3y-xright)left(y+1right)}sqrt{3y-2}-sqrt{dfrac{x+5}{2}}xy-2y-2end{matrix}right.
2. left{{}begin{matrix}sqrt{2y^2-7y+10-xleft(y+3right)}+sqrt{y+1}x+1sqrt{y+1}+dfrac{3}{x+1}x+2yend{matrix}right.
3. left{{}begin{matrix}sqrt{4x-y}-sqrt{3y-4x}12sqrt{3y-4x}+yleft(5x-yright)xleft(4x+yright)-1end{matrix}right.
4. left{{}begin{matrix}9sqrt{dfrac{41}{2}left(x^2+dfrac{1}{2x+y}right)}3+40xx^2+5xy+6y4y^2+9x+9end{matrix}right.
5. left{{}begin{mat...
Đọc tiếp
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
Giải hệa) left{{}begin{matrix}x^2left(y^2+1right)+2yleft(x^2+x+1right)3left(x^2+xright)left(y^2+yright)1end{matrix}right.b) left{{}begin{matrix}left(6x+5right)sqrt{2x+1}-2y-3y^30y+sqrt{x}sqrt{2x^2+4x-23}end{matrix}right.Giải bất ptdfrac{9}{left|x-5right|-3}geleft|x-2right|
Đọc tiếp
Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)
1)begin{cases}x^2-yleft(x+yright)+10left(x^2+1right)left(x+y-2right)+y0end{cases}2)begin{cases}x^2-4x+y^4+4y^22xy^2+2y^2+6x23end{cases}3)begin{cases}2x+frac{1}{x+y}34x^2+4y^2+4xy+frac{3}{left(x+yright)^2}7end{cases}4)begin{cases}y^6+x^9+3y^4+3y^284y^2-3x^3y^2+x^32end{cases}5)begin{cases}sqrt{x+y}-2sqrt{x-y}1x+sqrt{x^2+y^2}8end{cases}6) begin{cases}x+y-2frac{y}{x^2+1}x^2+y^2+xyy-1end{cases}7) begin{cases}4x-1sqrt{left(2x+yright).left(2y+1right)}sqrt{x+2y+1}-sqrt{x+y-1}sqrt{x-1}end{cases}8) begin{...
Đọc tiếp
1)\(\begin{cases}x^2-y\left(x+y\right)+1=0\\\left(x^2+1\right)\left(x+y-2\right)+y=0\end{cases}\)
2)\(\begin{cases}x^2-4x+y^4+4y^2=2\\xy^2+2y^2+6x=23\end{cases}\)
3)\(\begin{cases}2x+\frac{1}{x+y}=3\\4x^2+4y^2+4xy+\frac{3}{\left(x+y\right)^2}=7\end{cases}\)
4)\(\begin{cases}y^6+x^9+3y^4+3y^2=8\\4y^2-3x^3y^2+x^3=2\end{cases}\)
5)\(\begin{cases}\sqrt{x+y}-2\sqrt{x-y}=1\\x+\sqrt{x^2+y^2}=8\end{cases}\)
6) \(\begin{cases}x+y-2=\frac{y}{x^2+1}\\x^2+y^2+xy=y-1\end{cases}\)
7) \(\begin{cases}4x-1=\sqrt{\left(2x+y\right).\left(2y+1\right)}\\\sqrt{x+2y+1}-\sqrt{x+y-1}=\sqrt{x-1}\end{cases}\)
8) \(\begin{cases}\left(x+y\right).\left(x+4y^2+y\right)+3y^4=0\\\sqrt{x+2y^2+1}-y^2+y+1=0\end{cases}\)
ôi trờiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Đúng 0
Bình luận (0)
Giải hệ phương trình\(\left\{{}\begin{matrix}x-3y-2+\sqrt{y\left(x-y-1\right)+x}=0\\3\sqrt{8-x}-\dfrac{4y}{\sqrt{y+1}}+1=x^2-14y-8\end{matrix}\right.\)
\(\)\(\begin{cases}\left(\sqrt{2016+x^2}+x\right)\left(\sqrt{504+y^2}+y\right)=1008\\x\sqrt{6\text{x}-4\text{x}y+1}=8\text{x}y+6\text{x}+1\end{cases}\)\(\begin{cases}\sqrt{x+1}\left(x+5\right)-3\left(x-y+1\right)\sqrt{y}\left(y+4\right)=0\\x^2+6y+7=2\sqrt{3y+1}+3\sqrt{x+4y+5}\end{cases}\)
Xem chi tiết
bạn tách từng câu ra mik suy nghĩ từng câu
Đúng 0
Bình luận (1)
Giải hệ PT:
\(\hept{\begin{cases}x^2+4y-13+\left(x-3\right)\sqrt{x^2+y-4}=0\\\left(x+y-3\right)\sqrt{y}+\left(y-1\right)\sqrt{x+y+1}=x+3y-5\end{cases}}\)
Giải hệ phương trình:
1, left{{}begin{matrix}left(17-3xright)sqrt{5-x}+left(3y-14right)sqrt{4-y}02sqrt{2x+y+5}+3sqrt{3x+2y+11}x^2+6x+13end{matrix}right.
2, left{{}begin{matrix}xleft(x+yright)+sqrt{x+y}sqrt{2y}left(sqrt{2y^3}+1right)x^2y-5x^2+7left(x+yright)-46sqrt[3]{xy-x+1}end{matrix}right.
3, left{{}begin{matrix}sqrt{x}+sqrt[4]{32-x}-y^2+30sqrt[4]{x}+sqrt{32-x}+6y-240end{matrix}right.
Đọc tiếp
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x\left(x+y\right)+\sqrt{x+y}=\sqrt{2y}\left(\sqrt{2y^3}+1\right)\\x^2y-5x^2+7\left(x+y\right)-4=6\sqrt[3]{xy-x+1}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\end{matrix}\right.\)
1/PT (1) cho ta nhân tử x - y - 1:)
\(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\left(1\right)\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\left(2\right)\end{matrix}\right.\)
ĐK: \(x\le5;y\le4\); \(2x+y+5\ge0;3x+2y+11\ge0\)
PT (1) \(\Leftrightarrow\left(17-3x\right)\left(\sqrt{5-x}-\sqrt{4-y}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(3x-17\right)\left(\frac{x-y-1}{\sqrt{5-x}+\sqrt{4-y}}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(\frac{3x-17}{\sqrt{5-x}+\sqrt{4-y}}-3\sqrt{4-y}\right)=0\)
Dễ thấy cái ngoặc to < 0
Do đó x= y + 1
Thay xuống PT (2):\(y^2+8y+20=2\sqrt{3y+7}+3\sqrt{5y+14}\)\(\left(y+1\right)\left(y+2\right)=y^2+3y+2\)
ĐK: \(y\ge-\frac{7}{3}\) (để các căn thức được thỏa mãn)
PT (2) \(\Leftrightarrow y^2+3y+2+2\left(y+3-\sqrt{3y+7}\right)+3\left(y+4-\sqrt{5y+14}\right)=0\)
\(\Leftrightarrow\left(y^2+3y+2\right)\left(1+\frac{2}{y+3+\sqrt{3y+7}}+\frac{3}{y+4+\sqrt{5y+14}}\right)=0\)
Cái ngoặc to > 0 =>...
P/s: Is that true? Ko đúng thì chịu thua-_- Mất nửa tiếng đồng hồ để gõ bài này đấy:(
2/ĐK: \(x\ge-y;y\ge0\)
PT (1) \(\Leftrightarrow x\left(x+y\right)+\sqrt{x+y}=2y^2+\sqrt{2y}\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\sqrt{x+y}-\sqrt{2y}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}\right)=0\)
Cái ngoặc to \(\ge y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}>0\).
Do đó x = y \(\ge0\)
Thay xuống pt dưới: \(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)
Lập phương hai vế lên ra pt bậc 6, tuy nhiên cứ yên tâm, nghiệm rất đẹp: x = 1:)
Em đưa kết quả luôn: \(\left(x-1\right)\left(x^2-4x+7\right)\left(x^6-10x^5+56x^4-160x^3+272x^2-64x+40\right)=0\)
P/s: khúc cuối em ko còn cách nào khác nên đành lập phương:((
cho cac so thuc x va y thoa man
\(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)=1\)1
chung minh x+y=0
\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{ }x+1}=\frac{x+5}{2}\)\(\frac{x+5}{2}\)
Đúng 0
Bình luận (0)
Xét \(x^2+\sqrt{1+x^2}\)ta có:
\(x^2\ge0\)
nên \(1+x^2\ge1\)
\(\Rightarrow\sqrt{1+x^2}\ge\sqrt{1}=1\)
\(\Rightarrow x^2+\sqrt{1+x^2}\ge1\)
Tương tự ta có:
\(y^2+\sqrt{1+y^2}\ge1\)
Do đó: \(\left(x^2+\sqrt{1+x^2}\right)\left(y^2+\sqrt{1+y^2}\right)\ge1\)
Dấu bằng xảy ra khi \(x=0;y=0\)
Khi đó \(x+y=0\left(ĐPCM\right)\)
Đúng 0
Bình luận (0)
\(\left\{{}\begin{matrix}x^3-y^3+3y^2-3x-2=0\\x^2+\sqrt{1-x^2}-3\sqrt{2y-y^2}-1=0\end{matrix}\right.\)