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2 tháng 3 2020
e) \(⇔\left[\begin{array}{} x-1=0\\\ 2x+7=0\\ x^2+2=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=1\\\ x=-\frac{7}{2}\\ x^2=-2(ko.xảy.ra) \end{array}\right.\)\(⇔\left[\begin{array}{} x=1\\ x=-\frac{7}{2} \end{array}\right.\)
\(f) ⇔\left[\begin{array}{} 4x-10=0\\ 24+5x=0 \end{array}\right.\)\(⇔\left[\begin{array}{} x=\frac{10}{4}\\ x=-\frac{24}{5} \end{array}\right.\)
\(g) ⇔\left[\begin{array}{} 3,5-7x=0\\ 0,1x+2,3=0 \end{array}\right.⇔\left[\begin{array}{} x=0,5\\ x=-23 \end{array}\right.\)
\(h) ⇔\left[\begin{array}{} 5x+2=0\\ x-7=0 \end{array}\right.⇔\left[\begin{array}{} x=-\frac{2}{5}\\ x=7 \end{array}\right.\)
17 tháng 2 2020
\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)
4 tháng 5 2017
a) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}=\dfrac{5}{2}\\x=-\dfrac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{24}{5};\dfrac{5}{2}\right\}\)
b) \(\left(3.5-7x\right)\left(0.1x+2.3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3.5-7x=0\\0.1x+2.3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3.5}{7}=\dfrac{1}{2}\\x=-\dfrac{2.3}{0.1}=-23\end{matrix}\right.\)
Vậy \(S=\left\{-23;\dfrac{1}{2}\right\}\)
Y
12 tháng 1 2023
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
H
12 tháng 1 2023
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
CM
15 tháng 10 2017
(4x – 10)(24 + 5x) = 0 ⇔ 4x – 10 = 0 hoặc 24 + 5x = 0
4x – 10 = 0 ⇔ 4x = 10 ⇔ x = 2,5
24 + 5x = 0 ⇔ 5x = -24 ⇔ x = -4,8
Phương trình có nghiệm x = 2,5 và x = -4,8
PT
1
CM
10 tháng 4 2017
(3,5 – 7x)(0,1x + 2,3) = 0 ⇔ 3,5 – 7x = 0 hoặc 0,1x + 2,3 = 0
3,5 – 7x = 0 ⇔ 3,5 = 7x ⇔ x = 0,5
0,1x + 2,3 = 0 ⇔ 0,1x = - 2,3 ⇔ x = -23
Phương trình có nghiệm x = 0,5 hoặc x = -23
N
1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 3 2023
a.
\(x^2-x-\left(5x-5\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Câu b hoàn toàn tương tự
3 tháng 2 2022
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
a. (4x−10)(24+5x)=0⇔4x−10=0(4x−10)(24+5x)=0⇔4x−10=0 hoặc 24+5x=024+5x=0
+ 4x−10=0⇔4x=10⇔x=2,54x−10=0⇔4x=10⇔x=2,5
+ 24+5x=0⇔5x=24⇔x=−4,824+5x=0⇔5x=24⇔x=−4,8
Phương trình có nghiệm x = 2,5 và x = -4,8
b. (3,5−7x)(0,1x+2,3)=0⇔3,5−7x=0(3,5−7x)(0,1x+2,3)=0⇔3,5−7x=0hoặc 0,1x+2,3=00,1x+2,3=0
+ 3,5−7x=0⇔3,5=7x⇔x=0,53,5−7x=0⇔3,5=7x⇔x=0,5
+ 0,1x+2,3=0⇔0,1x=−2,3⇔x=−230,1x+2,3=0⇔0,1x=−2,3⇔x=−23
Phương trình có nghiệm x =0,5 hoặc x = -23