Giải phương trình z 2 + 4 z + 9 = 0 .
A. z = − 2 − i 5 hoặc z = - 2 + i 5
B. z = 2 − i 5 hoặc z = 2 + i 5
C. z = 2 − i 5 hoặc z = 2 + i 5
D. z = − 2 − i 5 2 hoặc z = − 2 + i 5 2
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a,
\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)
d,
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
mà A\(\le0\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\) phải bằng 0 đê thỏa mãn điều kiện
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy....
b;c)I hệt câu a nên làm tương tự nhá
d)
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)
B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)
Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)
Vậy....
a) Ta có (3 - 2i)z + (4 + 5i) = 7 + 3i <=> (3 - 2i)z = 7 + 3i - 4 - 5i
<=> z = <=> z = 1. Vậy z = 1.
b) Ta có (1 + 3i)z - (2 + 5i) = (2 + i)z <=> (1 + 3i)z -(2 + i)z = (2 + 5i)
<=> (1 + 3i - 2 - i)z = 2 + 5i <=> (-1 + 2i)z = 2 + 5i
z =
Vậy z =
c) Ta có + (2 - 3i) = 5 - 2i <=>
= 5 - 2i - 2 + 3i
<=> z = (3 + i)(4 - 3i) <=> z = 12 + 3 + (-9 + 4)i <=> z = 15 -5i
a) (3 + 2i)z – (4 + 7i) = 2 – 5i
⇔(3+2i)z=6+2i
<=> z = \(\dfrac{\text{6 + 2 i}}{\text{3 + 2 i}}\) = \(\dfrac{22}{13}\) - \(\dfrac{6}{13}\)i
b) (7 – 3i)z + (2 + 3i) = (5 – 4i)z
⇔(7−3i−5+4i)=−2−3i
⇔z= \(\dfrac{\text{− 2 − 3 i}}{\text{2 + i}}\) = \(\dfrac{-7}{5}\) - \(\dfrac{4}{5}i\)
c) z2 – 2z + 13 = 0
⇔ (z – 1)2 = -12 ⇔ z = 1 ± 2 √3 i
d) z4 – z2 – 6 = 0
⇔ (z2 – 3)(z2 + 2) = 0
⇔ z ∈ { √3, - √3, √2i, - √2i}
\(a,\) Sửa đề: \(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)
Ta thấy \(3x^2-12x+16=3\left(x-2\right)^2+4\ge4\Leftrightarrow\sqrt{3x^2-12x+16}\ge\sqrt{4}=2\)
\(y^2-4y+13=\left(y-2\right)^2+9\ge9\Leftrightarrow\sqrt{y^2-4y+13}\ge\sqrt{9}=3\)
Cộng vế theo vế 2 BĐT trên:
\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}\ge2+3=5\)
Dấu \("="\Leftrightarrow x=y=2\)
Vậy pt có nghiệm \(\left(x;y\right)=\left(2;2\right)\)
\(b,x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ \Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\\ \Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\\ \Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y-3=4\\z-5=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)
Đáp án A.