Giai pt : a, 3x+18=0
B, 6x-7=3x+2
C, 2x/x+3+4(x-3)/x =6
Giai toán hệ PT giúp m với
a. 3x^2+x-4=0
b, 2x^2-x-28=0
c. 6x^2-x-7=0
d. 3x^2-5=0
a) \(3x^2+x-4=0\)
\(\Leftrightarrow\)\(3x^2-3x+4x-4=0\)
\(\Leftrightarrow\)\(3x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-\frac{4}{3}\end{cases}}\)
Vậy..
b) \(2x^2-x-28=0\)
\(\Leftrightarrow\)\(\left(x-4\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-3.5\end{cases}}\)
Vậy...
c) \(6x^2-x-7=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{6}\end{cases}}\)
Vậy....
d) \(3x^2-5=0\)
\(\Leftrightarrow\)\(3x^2=5\)
\(\Leftrightarrow\)\(x^2=\frac{5}{3}\)
\(\Leftrightarrow\)\(x=\pm\sqrt{\frac{5}{3}}\)
Vậy...
giải các pt sau:
a, \(2x^3+6x^2+3x-18=0\)
b, \(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6\)
giải giúp mình đi
b. Bạn tham khảo tại đây
Câu hỏi của Võ Lan Nhi - Toán lớp 8 | Học trực tuyến
a) x+1/2x-6-4/2x-6 b) 3x-4/6x+3-x-5/6x+3
c) x-1/x-3-3x-8/3-x+3-2x/x-3 d) 3/x+5-5/x-7
e) 3/x+5-5/x-7 f) 2/x-2+3/x+2+5x-18/x2-4
Giai pt:
a. \(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
b. \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
Phân tích đa thức thành nhân tử
a) x³-3x²+3x-1-8y³
b) x⁴-4x³+8x²-16x+16
Giải pt
a) 6(x-3) +(x-1) ²-(x+1) ²=2x
b) (x+4) ²-(x+8) (x-8) =96
c) 4x²-1=(2x+1) (3x-5)
d) 2x²-x=3-6x
e) 2x³+5x²-3x=0
f) x(2x-7) -4x+14=0
g) (2x-5) ²-(x+2) ²=0
h) (3x+1) (7x+3) =(5x-7) (3x+1)
i) x²+10x+25-4x(x+5) =0
k))(4x-5) ²-2(16x²-25) =0
l) (4x+3) ²=4(x²-2x+1)
m) x²-11x+28=0
n) 3x³-3x²-6x=0
o) x²-9x+20=0
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(m,x^2-11x+28=0\)
\(\Leftrightarrow x^2-4x-7x+28=0\)
\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)
\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)
Tìm x biết :
a, 4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6(x + 14)
b, 5.(3x + 5) - 4.(2x - 3) = 5x + 3.(2x + 12) + 1
c, 2.(5x - 8) - 3.(4x - 5) = 4.(3x - 4) + 11
d, (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
e, (8x - 3)(3x + 2) - (4x + 7)(x + 4)= (2x + 1)(5x - 1) - 33
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1
=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
=> (15x - 8x) + (25 + 12) = 11x + 37
=> 7x + 37 = 11x + 37
=> 11x - 7x = 0
=> x = 0
tìm x biết
a) (6x-3) (2x+4) + (4x-1) (5-3x) = -21
b) 6x (3x+5) - 2x (9x-2) + (17-x) (x-1) + x (x-18) =0
c) (15-2x) (4x+1) - (13-4x) (2x-3) - (x-1) (x+2) + x2=52
d) (8x-3) (3x+2) - (4x+7) (x+4) = (2x+1) (5x-1) - 33
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21
<=> 12x2 + 24x - 6x - 12 + 20x - 12x2 - 5 + 3x = -21
<=> 41x = -21 + 12 + 5
<=> 41x = -4
<=> x = -4/41
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
B> <2X+5>2-<X+2>2=0
<2X+5-X-2><2X+X+2>=0
<X+3><3X+7>=0
X+3=0 HOẶC 3X+7=0
X=-3 HOẶC X=-7/3
C>X2-5X+6=0
X2-4X+4-X+2=0
<X-2>2-<X-2>=0
<X-2.><X-3>=0
X-2=0 HOẶC X-3=0
X=2 HOẶC X=3
D> <2X-7><2X-7-6<X-3>>=0
<2X-7><-4X+11>=0
2X-7=0 HOẶC -4X+11=0
X=7/2 HOẶC X=11/4
E><X-2><X+1>=X2-4
<X-2><X+1>-<X2-4>=0
<X-2><X+1>-<X-2><X+2>=0
-X+2=0
X=2
CÒN NHIÊU TỰ LÀM ĐI MỆT WA
Help me
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0