(3x+2)(2x-1)=(7x-6)(3x+2)
Tìm x:
1) -3.(1-2x) - 4.(1+3x) = -5x + 5
2) 3.(2x - 5) - 6.(1 - 4x) = -3x + 7
3) (1 - 3x) - 2.(3x - 6) = -4x - 5
4) x.(4x - 3) - 2x.(2x - 1) = 5x - 7
5) 3x.(2x - 1) - 6x.(x + 2) = -3x + 4
6) (1 - 2x).3 - 4.(6x - 1) = 7x - 5
7) 6x - 3.(1 - 4x) - 5.(x + 1) = 2x + 7
8) 6.(1 - 3x) - 3.(2x + 5) = -10x + 7
9) 3x.(1 - 2x) + 6x^2 - 7x = 8.(1 - 2x) - 9
10) 2x.(1 + 3x) - 3x.(4 + 2x) = 3x - 4
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
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
M(x)+N(x)=(3\(x^4\)-\(2x^3\)+\(5x^2-4x+1\))+(\(-3x^4+2x^3-3x^2+7x+5\))
=\(3x^4-2x^3+5x^2-4x+1-3x^4+2x^3-3x^2+7x+5\)
\(=(3x^4-3x^4)+(-2x^3+2x^3)+(5x^2-3x^2)+(-4x+7x)+(1+5) \)
\(=2x^2+3x+6\)
làm như trên có đúng k vậy
(1) 1>.-3x.(x2 +7x- 1/3)
2. -5x2y4(3x2y3-2x3y2-xy)
3. \(\dfrac{1}{2}\)x3y .(2x4y3-4xy-6)
4. -2x.(3x+20
5. 3x.(5-2x)
6. 2x.(2x2-4x+5)
7. 4x3y2.(-2x2y+4x4-3y2)
8.\(\dfrac{1}{2}\)x3y.( 2x4y3- 4xy -6)
1: \(=-3x^3-21x^2+x\)
2: \(=-15x^4y^7+10x^5y^6+5x^3y^5\)
3: \(=x^7y^4-2x^4y^2-3x^3y\)
5: \(=15x-6x^2\)
6: \(=4x^3-8x^2+10x\)
7: \(=-8x^5y^3+16x^7y^2-12x^3y^4\)
8: \(=x^7y^4-2x^4y^2-3x^3y\)
1) x2 - x - (3x - 3) = 02) x(x - 6) - 7x + 42 = 0
3) x3 - 3x2 + 3x - 1 = 0
4) (2x - 5)2 - (x + 2) 2 = 0
5) x(2x - 9) = 3x(x - 5)
1) x^2-x-(3x-3)=0
⇔ X^2-x-3x+3=0
⇔ x^2-4x+3 =0
⇔x^2-3x-x+3 =0
⇔ x(x-3)-(x-3) =0
⇔(x-1)(x-3) =0
⇔ x-1=0 -> x=1
x-3=0 -> x=3
Vậy tập nghiệm S={ 1;3}
/2x+3x/-3x+2=0
/2+3x/=4x-3
/7x+1/-/5x+6/=0
a) Ta có |2x + 3x| - 3x + 2 = 0
=> |2x + 3x| = 3x - 2
ĐK : 3x - 2 \(\ge0\Rightarrow x\ge\frac{2}{3}\)
Khi đó |2x + 3x| = 3x - 2
<=> \(\orbr{\begin{cases}2x+3x=3x-2\\2x+3x=-3x+2\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-2\\8x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)(loại)
Vậy không tìm được giá trị của x thỏa mãn
b) ĐK 4x - 3 \(\ge0\Rightarrow x\ge\frac{3}{4}\)
Khi đó |2 + 3x| = 4x - 3
<=> \(\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=\frac{1}{7}\left(\text{loại}\right)\end{cases}}\)
Vậy x = 5 là giá trị cần tìm
c) |7x + 1| - |5x + 6| = 0
=> |7x + 1| = |5x + 6|
=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{5}{2};-\frac{7}{12}\right\}\)là giá trị cần tìm
a) \(\left|2x+3x\right|-3x+2=0\)
<=> \(\left|5x\right|-3x+2=0\)
<=> \(\orbr{\begin{cases}5x-3x+2=0\left(x\ge0\right)\\-5x-3x+2=0\left(x< 0\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}\left(ktm\right)}\)
b) \(\left|2+3x\right|=4x-3\)
<=> \(\orbr{\begin{cases}2+3x=4x-3\left(x\ge-\frac{2}{3}\right)\\-2-3x=4x-3\left(x< -\frac{2}{3}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}3x-4x=-3-2\\-3x-4x=-3+2\end{cases}}\)
<=> \(\orbr{\begin{cases}-x=-5\\-7x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=\frac{1}{7}\left(ktm\right)\end{cases}}\)
c) \(\left|7x+1\right|-\left|5x+6\right|=0\)
<=> \(\left|7x+1\right|=\left|5x+6\right|\)
<=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)
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
tim x
-2x(6x-2)+3x(4x-7)=8
(7x-2)(2x+3)-(3x-5)(4x+6)=2x2-5
+(-2x+3)-6x(3x-4)=-22x2+7x
⇔(2 x ) 2+2.2 x .1+1 2=0 ... c ,(3 x −4) 2−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x 2−24 x +16−126 x 2−252 x +168 x +336+147 x +294=16.
https://olm.vn/hoi-dap/detail/192758180810.html
\(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)
<=> \(-12x^2+4x+12x^2-21x=8\)
<=> \(-17x=8\)
<=> \(x=-\frac{8}{17}\)
\(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)
<=> \(14x^2+17x-6-12x^2+2x+30=2x^2-5\)
<=> \(14x^2+17x-6-12x^2+2x+30-2x^2+5=0\)
<=> \(19x+29=0\)
<=> \(19x=-29\)
<=> \(x=-\frac{29}{19}\)
Ý cuối mình k biết -22x2 là -22.2 hay -22x2 nữa :)
x3y.( 2x4y3- 4xy -6)
1/2x^3y(2x^4y^3-4xy-6)
=1/2x^3y*2x^4y^3-1/2x^3y*4xy-1/2x^3y*6
=x^7y^4-2x^4y^2-3x^3y