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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
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
Có nghĩa khi
a/ \(2x-9\ge0\Rightarrow2x\ge9\Rightarrow x\ge\frac{9}{2}\)
b/ \(3-5x\ge0\Rightarrow-5x\ge-3\Rightarrow x\le\frac{3}{5}\)
c/ \(3x-3\ge0\Rightarrow3x\ge3\Rightarrow x\ge1\)
Đặt \(B=\frac{2\sqrt{x}+3}{\sqrt{x}-1}=\frac{2\sqrt{x}-2+5}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=2+\frac{5}{\sqrt{x}-1}\)
\(\Rightarrow B\in Z\Leftrightarrow2+\frac{5}{\sqrt{x}-1}\in Z\Leftrightarrow\frac{5}{\sqrt{x}-1}\in Z\Leftrightarrow5⋮\sqrt{x}-1\Leftrightarrow\sqrt{x}-1\inƯ\left(5\right)\)
\(\Rightarrow\sqrt{x}-1\in\left\{-5;-1;1;5\right\}\)
Vì x dương\(\Rightarrow\sqrt{x}-1\ge0\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;5\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2;6\right\}\)
\(\Rightarrow x\in\left\{4;36\right\}\)
Vậy số phần tử của tập hợp A là 2
\(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow\frac{x-1}{6}=\frac{x+5}{7}\)
\(\Leftrightarrow\frac{7\left(x-1\right)}{42}=\frac{6\left(x+5\right)}{42}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(x+5\right)\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow7x-6x=7+30\)
\(\Leftrightarrow x=37\)
Vậy nghiệm của phương trình là x = 37
a: \(\Leftrightarrow2\sqrt{x}=4\)
=>căn x=2
=>x=4