a) ||x|+3|-5=0
b) |2x-3|-7=0
c) ||x|-3|-7=0
Tìm x biết:
a) (2x - 3).(x + 5) = 0
b) 3x.(x - 2) - 7.(x - 2) = 0
c) 5x.(2x - 3) - 6x + 9 = 0
a)(2x-3)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy x=3/2 hoặc x=-5
a) \(\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)
b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)
c) \(5x\left(2x-3\right)-6x+9=0\)
\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)
a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)
c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Tìm x,biết:
a)(2x-3)2-49=0
b)2x.(x-5)-7.(5-x)=0
c)x2-3x-10=0
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Tìm các số thực x, biết:
a) (2x-3)2-49=0
b) 2x(x-5)-7(5-x)=0
c) x2-3x-10=0
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a. (2x - 3)2 - 49 = 0
<=> (2x - 3)2 - 72 = 0
<=> (2x - 3 + 7)(2x - 3 - 7) = 0
<=> (2x + 4)(2x - 10) = 0
<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b. 2x(x - 5) - 7(5 - x) = 0
<=> 2x(x - 5) + 7(x - 5) = 0
<=> (2x + 7)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
c. x2 - 3x - 10 = 0
<=> x2 - 5x + 2x - 10 = 0
<=> x(x - 5) + 2(x - 5) = 0
<=> (x + 2)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a, (2x - 3)2 - 49 = 0
(2x - 3)2 - 72 = 0
(2x - 3 + 7)( 2x - 3 - 7) = 0
(2x + 4)( 2x - 10) = 0
=> 2x + 4 = 0 => 2x - 10 = 0
2x = - 4 2x = 10
x = - 2 x = 5
Bài1:Giải phương trình:
a,(5-x)(3-2x)(3x+4)=0
b,(2x-1)(3x+2)(5-x)=0
c,(2x-1)(x-3)(x+7)=0
Giúp mình với :)
d,(3-2x)(6x+4)(5-8x)=0
a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)
a) (x-1/5).(8/5+2x)=0
b) (x-4/7):(x+1/2)>0
c) (2x-3):(x+7/4)<0
Mong mn trả lời ạ
\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
a) 4x – 20 = 0
b) 2x + x + 12 = 0
c) x – 5 = 3 – x
d) 7 – 3x = 9 – x
a) 4x – 20 = 0
⇔ 4x = 20
⇔ x = 20 : 4
⇔ x = 5
Vậy phương trình có nghiệm duy nhất x = 5.
b) 2x + x + 12 = 0
⇔ 3x + 12 = 0
⇔ 3x = -12
⇔ x = -12 : 3
⇔ x = -4
Vậy phương trình đã cho có nghiệm duy nhất x = -4
c) x – 5 = 3 – x
⇔ x + x = 5 + 3
⇔ 2x = 8
⇔ x = 8 : 2
⇔ x = 4
Vậy phương trình có nghiệm duy nhất x = 4
d) 7 – 3x = 9 – x
⇔ 7 – 9 = 3x – x
⇔ -2 = 2x
⇔ -2 : 2 = x
⇔ -1 = x
⇔ x = -1
Vậy phương trình có nghiệm duy nhất x = -1.
Em lớp 6 em chỉ làm dc phần a,b,c
Kết quả như sau:
a,4x-20=0
4x=20+0
4x=20
x=20:4
x=5
Tìm x biết:
a) 7x.(2x - 3) - (4x2 - 9) = 0
b) (2x - 7).(x - 2).(x2 - 4) = 0
c) (9x2 - 25) - (6x - 10) = 0
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
Tìm x biết:
a) (3x + 5)2 - 4x2 = 0
b) 25x4 - (4x - 3)2 = 0
c) (3x + 7)2 - (2x - 3)2 = 0
d) (4x - 1)2 - (5 - 3x)2 = 0
a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
a) 2x(x + 1) – 2x2 = 0
b) 3(x – 7) + 4x(x – 7) = 0
c) x2 – x = 12
\(a)2x(x+1)-2x^2=0 <=> 2x^2+2x-2x^2=0 \\<=>2x=0<=>x=0 \\b)3(x-7)+4x(x-7)=0<=>(4x+3)(x-7)=0 \\<=>4x+3=0\ hoặc\ x+7=0 \\<=>x=\dfrac{-3}{4}\ hoặc\ x=-7 \\c)x^2-x=12<=>x^2-x-12=0 \\<=>(x+3)(x-4)=0 \\<=>x+3=0\ hoặc\ x-4=0 \\<=>x=-3\ hoặc\ x=4\)