(x+11).(7-3x)=0
7)(16-8x)(2-6x)=0
8) (x+4)(6x-12)=0
9) (11-33x)(x+11)=0
10) (x-1/4)(x+5/6)=0
11) (7/8-2x)(3x+1/3)=0
12)3x-2x^2=0
7)(16-8x)(2-6x)=0
=> 16 - 8x = 0 hoặc 2 - 6x = 0
=> 16 = 8x hoặc 2 = 6x
=> x = 2 hoặc x = 1/3
8) (x+4)(6x-12)=0
=> x + 4 = 0 hoặc 6x - 12 = 0
=> x = -4 hoặc x = 2
9) (11-33x)(x+11)=0
=> 11 - 33x = 0 hoặc x + 11 = 0
=> x = 1/3 hoặc x = -11
10) (x-1/4)(x+5/6)=0
=> x - 1/4 = 0 hoặc x + 5/6 = 0
=> x = 1/4 hoặc x = -5/6
11) (7/8-2x)(3x+1/3)=0
=> 7/8 - 2x = 0 hoặc 3x + 1/3 = 0
=> 2x = 7/8 hoặc 3x = -1/3
=> x = 7/16 hoặc x = -1/9
12)3x-2x^2=0
=> x(3 - 2x) = 0
=> x = 0 hoặc 3 - 2x = 0
=> x = 0 hoặc x = 3/2
\(a,\left(16-8x\right)\left(2-6x\right)=0\)
\(\hept{\begin{cases}16-8x=0\\2-6x=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}}\)
\(b,\left(x+4\right)\left(6x-12\right)=0\)
\(\hept{\begin{cases}x+4=0\\6x-12=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\x=2\end{cases}}}\)
\(c,\left(11-33x\right)\left(x+11\right)=0\)
\(\hept{\begin{cases}11-33x=0\\x+11=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\x=-11\end{cases}}}\)
\(d,\left(x-\frac{1}{4}\right)\left(x+\frac{5}{6}\right)=0\)
\(\hept{\begin{cases}x-\frac{1}{4}=0\\x+\frac{5}{6}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{5}{6}\end{cases}}}\)
\(e,\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)
\(\hept{\begin{cases}\frac{7}{x}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{4}\\x=-\frac{1}{9}\end{cases}}}\)
\(f,3x-2x^2=0\)
\(x\left(3-2x\right)=0\)
\(\hept{\begin{cases}x=0\\3-2x=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Giải phương trình tích
7)(16-8x)(2-6x)=0
8) (x+4)(6x-12)=0
9) (11-33x)(x+11)=0
10) (x-1/4)(x+5/6)=0
11) (7/8-2x)(3x+1/3)=0
12)3x-2x^2=0
mk lưu nhầm ảnh ở bài dưới của câu
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
1).(4-3x)(10-5x)=0 2).(7-2x)(4+8x)=0 3).(9-7x)(11-3x)=0
4).(7-14x)(x-2)=0 5).(\(\dfrac{7}{8}\)-2x)(3x+\(\dfrac{1}{3}\))=0 6).3x-2x\(^2\)
7).5x+10x\(^2\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
Giải phương trình
1) 16-8x=0
2) 7x+14=0
3) 5-2x=0
4) 3x-5=7
5) 8-3x=6
6) 8=11x+6
7)-9+2x=0
8) 7x+2=0
9) 5x-6=6+2x
10) 10+2x=3x-7
11) 5x-3=16-8x
12)-7-5x=8+9x
13) 18-5x=7+3x
14) 9-7x=-4x+3
15) 11-11x=21-5x
16) 2(-7+3x)=5-(x+2)
17) 5(8+3x)+2(3x-8)=0
18) 3(2x-1)-3x+1=0
19)-4(x-3)=6x+(x-3)
20)-5-(x+3)=2-5x
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2
Vậy phương trình có nghiệm là x = 2
Giải phương trình tích
1) (4-3x)(10x-5)=0
2) (7-2x)(4+8x)=0
3) (9-7x)(11-3x)=0
4)(7-14x)(x-2)=0
5)(2x+1)(x-3)=0
6)(8-3x)(-3x+5)=0
phần cuối mk chụp ko đc hết . chỗ cuối là bằng \(\frac{-5}{-3}\)=\(\frac{5}{3}\)
a. (3x-1)x(-1/2x+5)=0
b. 2/3x+1/2x=5/2:15/4
c. (x.44/7+3/7).11/5-3/7=-2
d. 1/3x+2/5(x-1)=0
a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
gpt \(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)
\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)
\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)
\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)
Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)