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![](https://rs.olm.vn/images/avt/0.png?1311)
Từ định nghĩa bằng nhau của hai số phức, ta có:
a) ⇔
;
b) ⇔
;
c) ⇔
⇔
.
![](https://rs.olm.vn/images/avt/0.png?1311)
a)3x+yi=(2y+1)(2−x)i⇔{3x=2y+1y=2−x⇔{x=1y=13x+yi=(2y+1)(2−x)i⇔{3x=2y+1y=2−x⇔{x=1y=1
b)2x+y−1=(x+2y−5)i⇔{2x+y−1=0x+2y−5=0⇔{x=−1y=3
![](https://rs.olm.vn/images/avt/0.png?1311)
bài này mình chưa giải dc triệt để ở cái cuối
\(2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\) \(\left(y\le\dfrac{3}{2}\right)\)
\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)
\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)
\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)
\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)
\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)
\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)
\(\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right) ^2=x\sqrt{3-2y}-\sqrt{x}\)
\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)
\(\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\)
\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)
\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)
\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)
\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)
bình phương lên tính được u
\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)
Với \(x=0\) không phải nghiệm
Với \(x>0\) chia 2 vế cho pt đầu cho \(x^3\)
\(\Rightarrow2-\dfrac{4}{x}+\dfrac{3}{x^2}-\dfrac{1}{x^3}=2\left(2-y\right)\sqrt{3-2y}\)
\(\Leftrightarrow1-\dfrac{1}{x}+\left(1-\dfrac{1}{x}\right)^3=\sqrt{3-2y}+\sqrt{\left(3-2y\right)^3}\)
Xét hàm \(f\left(t\right)=t+t^3\Rightarrow f'\left(t\right)=1+3t^2>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow1-\dfrac{1}{x}=\sqrt{3-2y}\)
Thế vào pt dưới:
\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)-\sqrt{x}}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\left(\sqrt{x+1}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\sqrt{x+1}=2\)
Phương trình này ko có nghiệm đẹp, chắc bạn ghi nhầm đề bài của pt dưới
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\)
\(\Leftrightarrow x^2+2\le3x\)
Hoàn toàn tương tự ta có \(y^2+2\le3y\)
Do đó: \(P\ge\dfrac{x+2y}{3x+3y+3}+\dfrac{2x+y}{3x+3y+3}+\dfrac{1}{4\left(x+y-1\right)}\)
\(P\ge\dfrac{x+y}{x+y+1}+\dfrac{1}{4\left(x+y-1\right)}\)
Đặt \(a=x+y-1\Rightarrow1\le a\le3\)
\(\Rightarrow P\ge f\left(a\right)=\dfrac{a+1}{a+2}+\dfrac{1}{4a}\)
\(f'\left(a\right)=\dfrac{3a^2-4a-4}{4a^2\left(a+2\right)^2}=\dfrac{\left(a-2\right)\left(3a+2\right)}{4a^2\left(a+2\right)^2}=0\Rightarrow a=2\)
\(f\left(1\right)=\dfrac{11}{12}\) ; \(f\left(2\right)=\dfrac{7}{8}\) ; \(f\left(3\right)=\dfrac{53}{60}\)
\(\Rightarrow f\left(a\right)\ge\dfrac{7}{8}\Rightarrow P_{min}=\dfrac{7}{8}\) khi \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)
\(\Rightarrow xy\le1\)
\(P=4x^2y^2+2x^3+2y^3+10xy\)
\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)
Đặt \(xy=t\Rightarrow0< t\le1\)
Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)
\(\Rightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)
=>x=1/2;z=1/2;y=-3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2y^3-6y^2+7y-3=-2x\sqrt{1-x}+2\sqrt{1-x}+\sqrt{1-x}\)
\(\Leftrightarrow2\left(y^3-3y^2+3y+1\right)+y-1=2\left(1-x\right)\sqrt{1-x}+\sqrt{1-x}\)
\(\Leftrightarrow2\left(y-1\right)^3+y-1=2\left(\sqrt{1-x}\right)^3+\sqrt{1-x}\) (1)
Xét hàm \(f\left(t\right)=2t^3+t\)
\(f'\left(t\right)=6t^2+1>0\Rightarrow f\left(t\right)\) đồng biến
Nên (1) tương đương: \(y-1=\sqrt{1-x}\Rightarrow y=1+\sqrt{1-x}\)
\(\Rightarrow P=x+2\sqrt{1-x}+2=-\left(1-x-2\sqrt{1-x}+1\right)+4=-\left(\sqrt{1-x}-1\right)^2+4\le4\)
⇒ P = x + 2 √ 1 − x + 2
= − ( 1 − x − 2 √ 1 − x + 1 ) + 4
= − ( √ 1 − x − 1 ) 2 + 4 ≤ 4
Cho xin một like đi các dân chơi à.