Cho pt: x^2-(m+2)x+(2m-1)=0 có 2 nghệm x1,x2. Lập hệ thức lên hệ giữa x1,x2 không phụ thuộc m.
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\(pt\Leftrightarrow\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)^3=\left(\sqrt[3]{3x+2}-2\right)^3\)
\(\Leftrightarrow x-5+2x-1+3\sqrt[3]{x-5}\sqrt[3]{2x-1}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)=3x+2-8-3.2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow\sqrt[3]{2x^2-11x+5}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)=-2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\text{ (*)}\)
Mà: \(\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\)
\(+\text{Nếu }\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2=0\Leftrightarrow x=2\) thì pt thành 0 = 0 (thỏa mãn)
\(+\text{Xét }\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\ne0\)
\(\text{(*)}\Leftrightarrow\sqrt[3]{2x^2-11x+5}=-2\sqrt[3]{3x+2}\)
\(\Leftrightarrow2x^2-11x+5=-8\left(3x+2\right)\)
\(\Leftrightarrow2x^2+13x+21=0\)
\(\Leftrightarrow x=-3\text{ hoặc }x=-\frac{7}{2}\)
Vậy, tập nghiệm của phương trình là: \(S=\left\{-\frac{7}{2};\text{ }-3;\text{ }2\right\}\)
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a. \(9x=225\Rightarrow x=25\)
b. \(2x=8\Rightarrow x=4\)
c. \(3\left(2x-3\right)=6\Rightarrow6x=15\Rightarrow x=\frac{15}{6}\)
d. \(4\left(x+1\right)=8\Rightarrow4x=4\Rightarrow x=1\)
e. \(\sqrt{x+2}.\sqrt{x-2}-\sqrt{x-2}=0\Rightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)
hoặc \(\sqrt{x+2}-1=0\Rightarrow\sqrt{x+2}=1\Rightarrow x+2=1\Rightarrow x=-1\)
f. \(\sqrt{x+1}+3\sqrt{x+1}=4\Rightarrow4\sqrt{x+1}=4\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)
g. \(\sqrt{x-2}\left(1-\sqrt{x}\right)=0\)
=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)
hoặc \(1-\sqrt{x}=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
h. \(\sqrt{x+3}.\sqrt{x-3}-\sqrt{x+3}=0\Rightarrow\sqrt{x+3}\left(\sqrt{x-3}-1\right)=0\)
=> \(\sqrt{x+3}=0\Rightarrow x=-3\)
hoặc \(\sqrt{x-3}-1=0\Rightarrow\sqrt{x-3}=1\Rightarrow x=4\)
√x−2(1−√x)=0
=> √x−2=0⇒x−2=0⇒x=2
hoặc 1−√x=0⇒√x=1⇒x=1
h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0
=> √x+3=0⇒x=−3
hoặc
√x−2(1−√x)=0
=> √x−2=0⇒x−2=0⇒x=2
hoặc 1−√x=0⇒√x=1⇒x=1
h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0
=> √x+3=0⇒x=−3
hoặc
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\(a^4+b^4+\left(a+b\right)^2=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2+2ab\right)^2\)
\(=\left(a^2+b^2\right)-2a^2b^2+\left(a^2+b^2\right)+4ab\left(a^2+b^2\right)+4a^2b^2\)
\(=2\left[\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right)+a^2b^2\right]\)
\(=2\left(a^2+b^2+ab\right)^2\)
Tương tự: \(x^4+y^4+\left(x+y\right)^4=2\left(x^2+y^2+xy\right)^2\)
Mà \(a^2+b^2+\left(a+b\right)^2=x^2+y^2+\left(x+y\right)^2\Rightarrow2\left(a^2+b^2+ab\right)=2\left(x^2+y^2+xy\right)\)
\(\Rightarrow2\left(a^2+b^2+ab\right)^2=2\left(x^2+y^2+xy\right)^2\)
hay \(a^4+b^4+\left(a+b\right)^4=x^4+y^4+\left(x+y\right)^4\)