Tìm x \(\in\)\(ℤ\), biết:
\(\frac{4}{x+1}\)= \(\frac{2}{3x+1}\)
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7 tháng 9 2020
+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)
\(\Leftrightarrow3\sqrt{3x}=6\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)
Vậy \(S=\left\{\frac{4}{3}\right\}\)
+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{1,15\right\}\)
+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
\(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)
\(\Rightarrow\)\(\sqrt{x}-4< 0\)
\(\Leftrightarrow\)\(\sqrt{x}< 4\)
\(\Leftrightarrow\)\(x< 16\)
Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)
Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)
7 tháng 9 2020
\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) (Đk: x \(\ge\)0)
<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)
<=> \(3\sqrt{3x}=6\)
<=> \(\sqrt{3x}=2\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)
<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)
\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)
<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
<=> \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)
<=> \(\sqrt{x}< 4\) <=> \(x< 16\)
Kết hợp với đk => S = {x|0 < x < 16}
14 tháng 7 2021
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
HQ
29
3 tháng 12 2017
a, (x-15):5+22=24
( x - 15 ) : 5 = 2
x-15 = 10
x = 25
\(\frac{4}{x+1}=\frac{2}{3x+1}\Leftrightarrow4\left(3x+1\right)=2\left(x+1\right)\Leftrightarrow12x+4=2x+2\)
\(\Leftrightarrow12x-2x=2-4\Leftrightarrow10x=-2\Leftrightarrow\frac{-1}{5}\)
Vậy x=-1/5
\(\frac{4}{x+1}=\frac{2}{3x+1}\left(x\ne-1;x\ne-\frac{1}{3}\right)\)
=> \(4\left(3x+1\right)=2\left(x+1\right)\)
=> \(12x+4=2x+2\)
=> \(12x-2x=2-4\)
=> \(10x=-2\)
=> \(5x=-1\)(chia cho 5)
=> \(x=-\frac{1}{5}\left(tm\right)\)
Vậy \(x=-\frac{1}{5}\)