(x-5).(x-7)=0
Tìm x
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`(1/2x-7)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}\dfrac12x-7=0\\x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\dfrac12x=7\\x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=14\\x=-2\end{array} \right.\)
Vậy `x=14` hoặc `x=-2`
Ta có: \(\left(\dfrac{1}{2}x-7\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=14\\x=-2\end{matrix}\right.\)
a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)
a: (x-1)(x+2)(-x-3)=0
=>(x-1)(x+2)(x+3)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
b: (x-7)(x+3)<0
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
\(\Leftrightarrow2x^2-11x+5-2x^2+10x=25\Leftrightarrow-x=20\Leftrightarrow x=-20\)
PT có 2 nghiệm \(\Leftrightarrow\Delta'=\left(k-2\right)^2-\left(-2k-5\right)\ge0\)
\(\Leftrightarrow k^2-4k+4+2k+10\ge0\\ \Leftrightarrow k^2-2k+14\ge0\\ \Leftrightarrow k\in R\)
Vậy PT luôn có 2 nghiệm
Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=2\left(k-2\right)\left(1\right)\\x_1x_2=-2k-5\left(2\right)\end{matrix}\right.\)
Lại có \(2x_1-x_2=7\left(3\right)\)
\(\left(1\right)\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2\left(k-2\right)\\2x_1-x_2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x_1=2k+3\\x_2=2x_1-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{2k+3}{2}\\x_2=\dfrac{4k+6}{2}-7=\dfrac{4k-8}{2}=2k-4\end{matrix}\right.\)
Thay vào \(\left(2\right)\Leftrightarrow\dfrac{\left(2k+3\right)\left(2k-4\right)}{2}=-2k-5\)
\(\Leftrightarrow\left(2k+3\right)\left(k-2\right)=-2k-5\\ \Leftrightarrow2k^2-k-6+2k+5=0\\ \Leftrightarrow2k^2+k-1=0\\ \Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-1\end{matrix}\right.\)
\(\Leftrightarrow\left(x-2021\right)\left(x-5\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\)
\(\dfrac{1}{M}=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}-\dfrac{\sqrt{x}}{27}=\dfrac{27\sqrt{x}+54-x-5\sqrt{x}}{27\left(\sqrt{x}+5\right)}\)\(=\dfrac{-x+22\sqrt{x}+54}{27\left(\sqrt{x}+5\right)}\)
\(\Rightarrow\sqrt{x}.27B+135B=-x+22\sqrt{x}+54\)
\(\Leftrightarrow x+\sqrt{x}\left(27B-22\right)+135B-54=0\) (1)
Coi PT (1) là phương trình bậc 2 ẩn \(\sqrt{x}\)
PT (1) có nghiệm không âm \(\Leftrightarrow\left\{{}\begin{matrix}\Delta=729B^2-1728B+700\ge0\\S=22-27B\ge0\\P=135B-54\ge0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2}{5}\le B\le\dfrac{14}{27}\)
Suy ra \(max_B=\dfrac{14}{27}\Leftrightarrow x=16\)
A làm tương tự
\(\left(x-5\right).\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+5\\x=0+7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}\)
( x - 5 ) . ( x - 7 ) = 0
= > x lớn hơn 5 và - 7 = 0
Có : ( 7 - 5 ) . ( 7 - 7 ) = 0
Because : 2 . 0 = 0