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Question

giai pt

a.$x^{2}+\sqrt{x+1}=1$

$pt so phuc:(z +3i)(z^{2}-2z+5)=0$

score 2 · Answer 1

$(z+3i)(z^2-2z+5)=0$

$\Leftrightarrow (z+3i)[(z-1)^2+4]=0$

+ $z=-3i$

+ $(z-1)^2=-4=4i^2 \Rightarrow z-1 =\pm 2i \Rightarrow z=1\pm 2i$

Câu 1: Đặt $\sqrt{x+1}= t \ge 0 \Rightarrow t^2=x+1 \ (1)$ theo bài ra ta có $x^2 +t=1 \ (2)$

Từ $(1);\ (2)$ có hệ $\begin{cases} t^2 -x=1 \\ x^2 +t =1 \end{cases}$ trừ 2pt cho nhau được $(x-t)(x+t)+(x+t)=0$

+ $x=-t =-\sqrt{x+1} \Rightarrow x=\dfrac{1}{2}(1-\sqrt 5) $

+ $x-t+1=0 \Rightarrow \sqrt{x+1}=x+1 \Rightarrow x= -1;\ x=0$

score 1 · Answer 2

a. ĐK: $x\ge-1$

Ta có:

$x^2+\sqrt{x+1}=1$

$\Leftrightarrow \left\{\begin{array}{l}1-x^2\ge0\\x+1=(1-x^2)^2\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}1-x^2\ge0\\x^4-2x^2-x=0\end{array}\right.$

$\Leftrightarrow \left\{\begin{array}{l}1-x^2\ge0\\x(x+1)(x^2-x-1)=0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x=-1\\x=0\\x=\dfrac{1-\sqrt5}{2}\end{array}\right.$

2 Đáp án