1 tìm x biết
a;1/2+x^2-3/4=0
b;5/3/4-|2x+1/3|=(-1/2)^2
2timf a b c biết
2a=3b=5cva 2a -3b+c=6
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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{1}{2}=\dfrac{x}{-4}\\ \Rightarrow x=\dfrac{1.\left(-4\right)}{2}=\dfrac{-4}{2}=-2\)
a) 7/x-1=x+1/9
7/x-x-1=1/9
7/x-x=1/9+1
7/x-x=10/9
còn lại thì bạn tự tính nhé
câu b làm tương tự
a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)
\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)
\(\Leftrightarrow x=-11\)
\(a,\sqrt{x+1}< 2\Leftrightarrow\left\{{}\begin{matrix}x+1< 4\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x\ge-1\end{matrix}\right.\\ \Leftrightarrow-1\le x< 3\)
\(d,\sqrt{2x+1}\ge3\Leftrightarrow2x+1\ge9\Leftrightarrow x\ge4\)
a) `4\sqrt(2x-1)>8`
`<=>\sqrt(2x-1)>2`
`<=>2x-1>4`
`<=>x>5/2`
b) `2\sqrtx-1>3`
`<=>2\sqrtx>4`
`<=>\sqrtx>2`
`<=>x>4`
a) Ta có: \(4\sqrt{2x-1}>8\)
\(\Leftrightarrow2x-1>4\)
\(\Leftrightarrow2x>5\)
hay \(x>\dfrac{5}{2}\)
b) Ta có: \(2\sqrt{x}-1>3\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4