(3x+2)-(x-1)=4(x+1)
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30 tháng 7 2021
1) Ta có: \(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1-21+x=0\)
\(\Leftrightarrow x^2-3x-20=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)
30 tháng 7 2021
1)\(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow21-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
2)\(\sqrt{8-x}+2=x\)
\(\Leftrightarrow8-x=\left(x-2\right)^2\)
\(\Leftrightarrow8-x=x^2-4x+4\)
\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
KK
2 tháng 6 2023
D = (3x - 2)^2 - 3(x - 4)(4 + x) + (x - 3)^3 - (x^2 - x + 1)(x + 1)
D = 9x^2 - 12x + 4 - 3x^2 + 48 + x^3 - 9x^2 + 27x - 27 - x^3 - 1
D = -3x^2 + 15x + 24
T
20 tháng 1 2024
a. ĐKXĐ: \(x\ne1\)
\(\dfrac{11x-4}{x-1}+\dfrac{10x+4}{2-2x}=\dfrac{2\cdot\left(11x-4\right)}{2\cdot\left(x-1\right)}-\dfrac{10x+4}{2x-2}\)
\(=\dfrac{22x-8}{2\left(x-1\right)}-\dfrac{10x+4}{2\left(x-1\right)}\)\(=\dfrac{22x-8-10x-4}{2\left(x-1\right)}\)
\(=\dfrac{12x-12}{2\left(x-1\right)}\)\(=\dfrac{12\left(x-1\right)}{2\left(x-1\right)}=6\)
b. ĐKXĐ: \(x\ne-2;x\ne\dfrac{1}{2}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}+\dfrac{5}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2x-1+5}{\left(2x-1\right)\left(x+2\right)}=\dfrac{2x+4}{\left(2x-1\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)}{\left(2x-1\right)\left(x+2\right)}=\dfrac{2}{2x-1}\)
\(\text{#}Toru\)
NQ
1
10 tháng 3 2020
(3x-2) (9x+6x+4)-(3x-1) (9x+3x+1)=x-4
(3x - 2)(15x + 4) - (3x - 1)(12x + 1) = x - 4
<=> 45x2 + 12x - 30x - 8 - (36x2 + 3x - 12x - 1) - x + 4 = 0
<=> 9x2 - 10x - 3 = 0
<=> (3x - \(\frac{5}{3}\))2 = \(\frac{52}{9}\) => \(\orbr{\begin{cases}3x-\frac{5}{3}=\frac{2\sqrt{13}}{3}\\3x-\frac{5}{3}=-\frac{2\sqrt{13}}{3}\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{5+2\sqrt{13}}{9}\\x=\frac{5-2\sqrt{13}}{9}\end{cases}}\)
Vậy ...
1G
1
25 tháng 7 2018
\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=4\)
\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)
\(\Rightarrow2x+4=4\)
\(\Rightarrow2x=0\Leftrightarrow x=0\)
14 tháng 1 2022
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
14 tháng 1 2022
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
3x+2-x+1=4x+4
2x+3=4x+4
2x-4x=4-3
-2x=1
x=-1/2
tick cho mik nhe