Giai phuong trinh:
\(\frac{15}{2}\left(30x^2-4x\right)=2004\left(\sqrt{30060x+1}+1\right)\)
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Giải phương trình :
1)\(\sqrt[3]{6x+1}=8x^3-4x-1\\ \)
2)\(2x+\frac{x-1}{x}=\sqrt[2]{1-\frac{1}{x}}+3\sqrt{x-\frac{1}{x}}\)
3)\(\frac{15}{2}\left(30x^2-4x\right)=2004\left(\sqrt{30060x+1}+1\right)\)
Ai biết giải giúp mình nha,mình nổ não lun mà ko giải được.
GIAI PHUONG TRINH:
\(2\left(\sqrt{\frac{x-1}{4}-3}\right)=2\sqrt{\frac{4x-4}{9}}-\frac{1}{3}\)
giai phuong trinh\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x-2-5x-5=15\)
\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)
Vậy \(S=\left\{\frac{-11}{2}\right\}\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow-4x-7=15\)
\(\Leftrightarrow-4x=22\)
\(\Leftrightarrow x=22:\left(-4\right)\)
\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)
Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)
\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\) giai phuong trinh
Giai phuong trinh sau :
c ) \(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)
\(\left(2-x\right)\left(2x-1\right)+\left(4x^2-4x+1\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(2x-1\right)+\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2-x+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-1\end{array}\right.\)
Vậy phương trình có tập nghiệm \(\left\{-1;\frac{1}{2}\right\}\)
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(2-x)(2x-1)+(4x^2-4x+1)=0
Ta có: (2x-1)(2-x)+(2x-1)^2=0
(2x-1)(2-x+2x-1)=0
Sau đó bn tự lam nha tại vì mk làm bằng phone
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Giai phuong trinh
\(\sqrt{x.\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2x\)
x=0
nhanh nhất có thể
X=0
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Giai phuong trinh:
\(\sqrt{x+\frac{3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\)
\(DKXD:x>0\)
\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(\frac{1}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{1}{2\left(x+1\right)}\right)=0\)
\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\left(\text{ }x-1\right)\left(x^2+x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy PT có 2 nghiệm \(x=1;x=3\)
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Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}x^2-y^2=1\\4x^2-5xy=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+\sqrt{y+2018}=1\\\sqrt{x+2018}+y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x+y=\sqrt{4z-1}\\y+z=\sqrt{4x-1}\\z+x=\sqrt{4y-1}\end{matrix}\right.\)
Giai phuong trinh giup minh voi:
\(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x-1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)