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\(\left(0.25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)
\(=0.4^{10}+\sqrt{25-9}\)
\(=0+\sqrt{16}=0+4=4\)
\(\dfrac{5}{20}+\dfrac{18}{11}-25\%-\left(\dfrac{18}{11}-\dfrac{4}{9}\right)\)
\(=\dfrac{5}{20}+\dfrac{18}{11}-\dfrac{1}{4}-\dfrac{18}{11}+\dfrac{4}{9}\)
\(=\left(\dfrac{5}{20}-\dfrac{1}{4}\right)+\left(\dfrac{18}{11}-\dfrac{18}{11}\right)+\dfrac{4}{9}\)
\(=0+0+\dfrac{4}{9}=\dfrac{4}{9}\)
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(3√x+5-5).2-20=18
=>(3\(\sqrt{x}\)).2=18+20
=>(3\(\sqrt{x}\)).2=38
=>(3\(\sqrt{x}\))=38:2
=>3\(\sqrt{x}\)=19
=>\(\sqrt{x}\)=19:3
=>\(\sqrt{x}\)=\(\frac{19}{3}\) (vô lí)
Vậy không có giá trị nào của x
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Sửa đề: \(5\left(1+\sqrt{1+x^3}\right)=x^2\left(4x^2-25x+18\right)\)
Đặt \(\sqrt{1+x^3}=a>0\)
Thì ta có:
\(5\left(1+a\right)=4x^4-25x^3+18x^2\)
\(\Leftrightarrow4x^4-25a^2+18x^2+20-5a=0\)
\(\Leftrightarrow\left(2x^2-5a+4\right)\left(2x^2+5a+5\right)=0\)
Với \(2x^2+4=5a\)
\(\Leftrightarrow2x^2+4=5\sqrt{1+x^3}\)
\(\Leftrightarrow\left(2x^2+4\right)^2=25\left(1+x^3\right)\)
\(\Leftrightarrow\left(x^2-5x-3\right)\left(4x^2-5x+3\right)=0\)
Tương tự cho trường hợp còn lại.
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a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
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1.
a) \(x-4\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=0+4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy \(x\in\left\{0;16\right\}.\)
b) \(\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|-\frac{3}{4}=\frac{1}{5}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{1}{5}+\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{19}{20}.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}-\frac{1}{20}=\frac{19}{20}\\\frac{3}{5}\sqrt{x}-\frac{1}{20}=-\frac{19}{20}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}=1\\\frac{3}{5}\sqrt{x}=-\frac{9}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1:\frac{3}{5}\\\sqrt{x}=\left(-\frac{9}{10}\right):\frac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{5}{3}\\\sqrt{x}=-\frac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{25}{9}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\frac{25}{9}.\)
Câu c) làm tương tự như câu b).
Chúc bạn học tốt!