K.(3x-1)(2x+7)-(x+1)(6x-5)=16
L.4x(x-7)=(x-7)
M.2(x+15)-x^2-5x=0
N.(x-2)^2-16=0
Những câu hỏi liên quan
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
\(m,x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
\(n,2x^2+5x-3=0\)
\(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
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\(k,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-4x-7x+14=0\)
\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
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Xem thêm câu trả lời
1) (4-3x) (10x-5)02) (7-2x) (4+8x) 03) (9-7x) (11-3x) 04) (7-14x) (x-2) 05) (2x+1) (x-3) 06) (8-3x) (-3x+5) 07) (16-8x) (2-6x) 08) (x+4) (6x-12) 09) (11-33x) (x+11) 010) (x-1/4) (x+5/6) 011) (7/8-2x) (3x+1/3) 012) 3x - 2x^2 013) 5x + 10x^2 014) 4x + 3x^2 015) -8x^2 + x 016) 10x^2 - 15x 017) x^2 -4 018) 9 - x^2 019) x^2 -1 020) (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)(...
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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
Tìm x, biết:
a)2x*(6x-5)-4x*(3x+7)=7
b)-5x(2x+1)+3x(3x+2)+x(x-1/2)=0
c)3x*(6x-5)-2x(9x+7)=15
d)1/2x*(2x+4)-(x+3)=5
e)(-3x+2)*5x-5x*(2x+1)-5x=4
Mấy bạn giúp mk ik mk đang cần gấp!
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
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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...
https://i.imgur.com/PCDykdb.jpg
1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2
Vậy phương trình có nghiệm là x = 2
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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
M) (2x+3)(-x+7)=0
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
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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
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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
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Xem thêm câu trả lời
Tìm x biết
1.(x+3)2-(x+2).(x-2)=4x+17
2.(2x+1)2-(4x-1).(x-3)-15=0
3.(2x+3).(x-1)+(2x-3).(1-x)=0
4.2(5x-8)-3(4x-5)=4(3x-4)+11
5.(3x-1).(2x-7)-(1-3x).(6x-5)=0
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
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Phân tích đa thức thành nhân tửa) x³-3x²+3x-1-8y³b) x⁴-4x³+8x²-16x+16Giải pta) 6(x-3) +(x-1) ²-(x+1) ²2xb) (x+4) ²-(x+8) (x-8) 96c) 4x²-1(2x+1) (3x-5) d) 2x²-x3-6xe) 2x³+5x²-3x0f) x(2x-7) -4x+140g) (2x-5) ²-(x+2) ²0h) (3x+1) (7x+3) (5x-7) (3x+1) i) x²+10x+25-4x(x+5) 0k))(4x-5) ²-2(16x²-25) 0l) (4x+3) ²4(x²-2x+1) m) x²-11x+280n) 3x³-3x²-6x0o) x²-9x+200
Đọc tiếp
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
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
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\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
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\(m,x^2-11x+28=0\)
\(\Leftrightarrow x^2-4x-7x+28=0\)
\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)
\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)
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Xem thêm câu trả lời
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
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2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
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