\(\left\{{}\begin{matrix}x_1^2-x_2^2=6\\x_1+x_2=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1-x_2\right)\left(x_1+x_2\right)=6\\x_1+x_2=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=-3\\x_1+x_2=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x_1=-5\\x_1+x_2=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{5}{2}\\-\dfrac{5}{2}+x_2=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1=-\dfrac{5}{2}\\x_2=\dfrac{1}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm: \(x_1=-\dfrac{5}{2};x_2=\dfrac{1}{2}\)

a) thay m=2 vào pt (1) ta có

\(x^2-3x+2=0\)

<=>\(\left(x-1\right)\left(x+2\right)=0\)

<=>\(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b)để pt (1) có 1 nghiệm

<=>\(\Delta=0\)

<=>9-4m=0

<=>m=\(\dfrac{9}{4}\)

KL: vậy để pt (1) có 1 nghiệm thì m=\(\dfrac{9}{4}\)

c)để pt có 2 nghiệm pb thì \(\Delta>0\)

<=>9-4m>0

<=>m<\(\dfrac{9}{4}\)

áp dụng định lý Vi-ét ta có \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=m\end{matrix}\right.\)

theo đề bài ta có \(x_1^3x_2+x_1x_2^3-2x_1^2x_2^2=5\)

<=>\(x_1x_2\left(x_1+x_2\right)^2-4x^2_1x^2_{2^{ }}=5\)

<=>\(9m-4m^2=5\)

<=>\(4m^2-9m+5=0\)

<=>\(\left(m-1\right)\left(4m-5\right)=0\)

<=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{5}{4}\end{matrix}\right.\)

KL: vậy với m =1 hoặc m=\(\dfrac{5}{4}\) thì pt có 2 nghiệm pb thỏa yêu cầu đề bài

P=\(x+y-\sqrt{xy}-x-y+2\sqrt{xy}\)=\(\sqrt{xy}\)

\(P=\dfrac{\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)

\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

\(=\sqrt{xy}\)

\(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{2}{x-1}\)

Vậy \(Q=\dfrac{2}{x-1}\)

Sinh khối là dạng vật liệu sinh học từ sự sống, hay gần đây là sinh vật sống, đa số là các cây trồng hay vật liệu có nguồn gốc từ thực vật . Được xem là nguồn năng lượng tái tạo, năng lượng sinh khối có thể dùng trực tiếp, gián tiếp một lần hay chuyển thành dạng năng lượng khác như nhiên liệu sinh học. Sinh khối có thể chuyển thành năng lượng theo ba cách: chuyển đổi nhiệt, chuyển đổi hóa học , và chuyển đổi sinh hóa .

Sinh khối là dạng vật liệu sinh học từ sự sống, hay gần đây là sinh vật sống, đa số là các cây trồng hay vật liệu có nguồn gốc từ thực vật. Được xem là nguồn năng lượng tái tạo, năng lượng sinh khối có thể dùng trực tiếp, gián tiếp một lần hay chuyển thành dạng năng lượng khác như nhiên liệu sinh học.

Sinh khối là dạng vật liệu sinh học từ sự sống, hay gần đây là sinh vật sống, đa số là các cây trồng hay vật liệu có nguồn gốc từ thực vật.Được xem là nguồn năng lượng tái tạo, năng lượng sinh khối có thể dùng trực tiếp, gián tiếp một lần hay chuyển thành dạng năng lượng khác như nhiên liệu sinh học. Sinh khối có thể chuyển thành năng lượng theo ba cách: chuyển đổi nhiệt, chuyển đổi hóa học, và chuyển đổi sinh hóa.

\(x^2+2x^2y^2+2y^2-\left(x^2y^2+2x^2\right)-2=0\)

\(\Leftrightarrow x^2+2x^2y^2+2y^2-x^2y^2-2x^2-2=0\)

\(\Leftrightarrow x^2y^2-x^2+2y^2-2=0\)

\(\Leftrightarrow\left(x^2y^2-x^2\right)+\left(2y^2-2\right)=0\)

\(\Leftrightarrow x^2\left(y^2-1\right)+2\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(y-1\right)\left(y+1\right)=0\)

Dễ thấy: \(x^2+2\ge2>0\forall x\) (vô nghiệm)

\(\Rightarrow\left[{}\begin{matrix}y-1=0\\y+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

x2 + 2x2y2 + 2y2 − (x2y2+2x2) − 2=0

x2+2x2y2+2y2−x2y2−2x2−2=0

Dễ thấy:  (không có nghiệm)

⇒y−1=0 hoặc y+1=0⇒𝑦−1=0 hoặc 𝑦+1=0⇒y=1 hoặc y=−1

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Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy: S={0;4}

\(x^2-4x=0\)

\(x\left(x-4\right)=0\)

\(x=0\) hoặc \(x-4=0\)

\(x=0\) hoặc \(x=4\)

Vậy..........

Nhớ tick và theo dõi mình nha!

x² - 4x = 0

=> x . x = 4 . x

=> x = 4