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1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)
Ta xét các trường hợp sau:
Trường hợp 1:
\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:
\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)
\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)
Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)
Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:
\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)
Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)
+ Nếu như:
\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)
+ Nếu như:
\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)
Trường hợp 2:
\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:
\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)
Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)
Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)
\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)
Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)
\(\Leftrightarrow2b=1-\frac{1}{a}\)
Thay vào (1) ta được :
\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)
Giải pt được \(a=1\)
Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)
Ta có hệ :
\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy...
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(2x^3-2y^3+5x-5y=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)
\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)
Thế vào pt dưới:
\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)
Đặt \(x+\frac{1}{x}+1=t\)
\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)
\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}+1=-1\\x+\frac{1}{x}+1=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-2x+1=0\end{matrix}\right.\) \(\Leftrightarrow...\)
a/ ĐKXĐ: ...
\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)
TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)
TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)
\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)
\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)
Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)
\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)
\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)
Vế trái luôn dương nên pt vô nghiệm
a)
HPT \(\Leftrightarrow \left\{\begin{matrix} 4x+8y=0(1)\\ 4x+2y=-3(2)\end{matrix}\right.\)
Lấy $(1)-(2)$ ta thu được: $8y-2y=3$
$\Leftrightarrow 6y=3\Leftrightarrow y=\frac{1}{2}$
Khi đó: $x=\frac{-4y}{2}=-2y=-1$
Vậy..........
b)
HPT \(\Leftrightarrow \left\{\begin{matrix} 2x-y=-4(1)\\ 2x+4y=-6(2)\end{matrix}\right.\)
Lấy $(1)-(2)$ suy ra: $-y-4y=-4-(-6)$
$\Leftrightarrow -5y=2\Rightarrow y=\frac{-2}{5}$
$\Rightarrow x=-3-2y=\frac{-11}{5}$
c)
HPT \(\Leftrightarrow \left\{\begin{matrix} xy+2x-15y-30=xy\\ xy-x+15y-15=xy\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x-15y=30\\ -x+15y=15\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 2x-15y=30(1)\\ -2x+30y=30(2)\end{matrix}\right.\)
Lấy $(1)+(2)$ suy ra $-15y+30y=60$
$\Leftrightarrow 15y=60\Leftrightarrow y=4$
$\Rightarrow x=15y-15=45$
Vậy.......
d)
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}+\frac{2}{y}=10(1)\\ \frac{2}{x}+\frac{5}{y}=7(2)\end{matrix}\right.\)
Lấy \((2)-(1)\Rightarrow \frac{3}{y}=7-10=-3\Rightarrow y=-1\)
\(\Rightarrow \frac{1}{x}=5-\frac{1}{y}=5-\frac{1}{-1}=6\Rightarrow x=\frac{1}{6}\)
Vậy........
a: ĐKXĐ: y<=1/2
\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6\left(x-1\right)-2\sqrt{1-2y}=2\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x-1\right)=7\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=1\\2\sqrt{1-2y}=5-1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\\sqrt{1-2y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\1-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
b:
ĐKXĐ: \(x\in R\)
\(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{\left(x-1\right)^2}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x-1\right|-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-6y=14\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=13\\\left|x-1\right|-3y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\\left|x-1\right|=3y+7=3\cdot\dfrac{13}{2}+7=\dfrac{39}{2}+7=\dfrac{53}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x-1\in\left\{\dfrac{53}{2};-\dfrac{53}{2}\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x\in\left\{\dfrac{55}{2};-\dfrac{51}{2}\right\}\end{matrix}\right.\)
c: ĐKXĐ: y>=4
\(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-x\right)+2\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-x\right)=-7\\2\left(x^2-x\right)+\sqrt{y-4}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x=-1\\\sqrt{y-4}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+1=0\\y-4=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vôlý\right)\\y=8\end{matrix}\right.\)
=>\(\left(x,y\right)\in\varnothing\)
a/\(\left(x+3\right)\left(\frac{40}{x}+3\right)=88\)(đk: x\(\ne0\))
\(\Leftrightarrow40+\frac{120}{x}+3x+9=88\)
\(\Leftrightarrow3x+\frac{120}{x}=39\) \(\Leftrightarrow\frac{3x^2+120}{x}=39\)
\(\Leftrightarrow x^2+40-13x=0\Leftrightarrow\left(x-8\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)(tm)
vậy x=8 hoặc x=5 là nghiệm của phương trình
b/ \(\frac{x-1}{2x+1}< \frac{1}{2}\left(x\ne-\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{x-1}{2x+1}-\frac{1}{2}< 0\) \(\Leftrightarrow\frac{2x-2}{4x+2}-\frac{2x+1}{4x+2}< 0\)
\(\Leftrightarrow\frac{-3}{4x+2}< 0\) \(\Leftrightarrow4x+2>0\)(vì -3<0)
<=> x>-1/2(tm)
vậy x>-1/2 là nghiệm của phương trình
c/ \(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=2y-2x\left(1\right)\\x^3+1=2y\left(2\right)\end{matrix}\right.\)
(1)\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
\(\Leftrightarrow x=y\)(vì x^2 +xy+y^2 +2>0)
thay x=y vào (2) ta được:
\(x^3-2x+1=0\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\sqrt{5}-1}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
vậy...