Gpt: \(6x^2+13x-5=0\)
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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
c.
ĐLXĐ: \(x\ge-\dfrac{1}{3}\)
\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)
Đặt \(\sqrt{3x+1}=t\ge0\)
\(\Rightarrow-t^2+t+4x^2-10x+6=0\)
\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 7 2021
a.
ĐKXĐ: \(x\ge-\dfrac{5}{4}\)
\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)
\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)
\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)
\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
AT
30 tháng 6 2017
a, \(x^2-x-14x+14=0\)
\(=>x\left(x-1\right)-14\left(x-1\right)=0\)
\(=>\left(x-14\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-14=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=14\\x=1\end{matrix}\right.\)
b, \(x^2+2x+7x+14=0\)
\(=>x\left(x+2\right)+7\left(x+2\right)=0\)
\(=>\left(x+7\right)\left(x+2\right)=0\)
\(< =>\left\{{}\begin{matrix}x+7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=-2\end{matrix}\right.\)
c, \(6x^2-6x-5x+5=0\)
\(=>6x\left(x-1\right)-5\left(x-1\right)=0\)
\(=>\left(6x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}\\x=1\end{matrix}\right.\)
d, \(6x^2+3x+10x+5=0\)
\(=>3x\left(2x+1\right)+5\left(2x+1\right)=0\)
\(=>\left(3x+5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
e, \(10x^2+10x+3x+3=0\)
\(=>10x\left(x+1\right)+3\left(x+1\right)=0\)
\(=>\left(10x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{10}\\x=-1\end{matrix}\right.\)
CHÚC BẠN HỌC TỐT...
16 tháng 1 2018
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
25 tháng 11 2016
a)\(6x^2+5x-6=0\)
\(\Leftrightarrow6x^2-4x+9x-6=0\)
\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
b)\(6x^2-13x+6=0\)
\(\Leftrightarrow6x^2-4x-9x+6=0\)
\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
c)\(10x^2-13x-3=0\)
\(\Leftrightarrow10x^2-15x+2x-3=0\)
\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)
d)\(20x^2+19x-3=0\)
\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)
\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)
e)\(3x^2-x+6=0\)
\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)
Suy ra vô nghiệm
\(6x^2+13x-5=0\)
\(\Leftrightarrow6x^2-2x+15x-5=0\)
\(\Leftrightarrow2x\left(3x-1\right)+5\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-5\\3x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{1}{3}\end{cases}}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{\frac{-5}{2};\frac{1}{3}\right\}\)