4x-10* x o x m n 3x-5*
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Những câu hỏi liên quan
Câu 1. Giải các phườn trình sau:
a, 3x+60
b, 2x-100
c, 3x-711
d, 3x-90
e, 3x(2-x) 15(x-2)
f, (x+5)(x+4)0
g, x(x+4)0
h, (2x -4)(x-2)0
i, (x+1/5)(2x-3)0
k, x²-4x0
m, 4x²-10
n, x²-6x+90
l, (3x-5)²-(x+4)²0
o, 7x(x+2)-5(x+2)0
p, 3x(2x-5)-4x+100
q, (2-2x)-x²+10
r, x(1-3x)5(1-3x)
s, 2x-3/4+x+1/63
t, x-3/4-2x+1/3x/6
u, x+1/13+x+2/12x+3/11+x+4/10
v, 2x+1/15+2x+2/142x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
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Câu 1. Giải các phườn trình sau:
a, 3x+6=0
b, 2x-10=0
c, 3x-7=11
d, 3x-9=0
e, 3x(2-x) =15(x-2)
f, (x+5)(x+4)=0
g, x(x+4)=0
h, (2x -4)(x-2)=0
i, (x+1/5)(2x-3)=0
k, x²-4x=0
m, 4x²-1=0
n, x²-6x+9=0
l, (3x-5)²-(x+4)²=0
o, 7x(x+2)-5(x+2)=0
p, 3x(2x-5)-4x+10=0
q, (2-2x)-x²+1=0
r, x(1-3x)=5(1-3x)
s, 2x-3/4+x+1/6=3
t, x-3/4-2x+1/3=x/6
u, x+1/13+x+2/12=x+3/11+x+4/10
v, 2x+1/15+2x+2/14=2x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
r, x(1-3x)=5(1-3x)
➜x(1-3x)-5(1-3x)=0
➜(x-5)(1-3x)=0
➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Mk lười lắm mai nha!!!~~~~~~~~~~~~
Làm dần:
a, 3x+6=0
➜3x=-6
➜x=2
b, 2x-10=0
➜2x=10
➜x=5
c, 3x-7=11
➜3x=11+7
➜3x=18
➜x=6
d, 3x-9=0
➜3x=9
➜x=3
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C1.10x2=6x+8
C2.23x+10=23+13x
C3.9x-6=4x+1
C4.15x-12=11x+15
C5.21x+9=19-11x
C6.15+16x=8-3x
C7.19-4x=8x+23
C8.51-10x=3x-21
C9.8-6x=11-4x
C10.2(3x+4)-3(1-2x)=8x+10
C11.5(3-4x)-4(2x-5)=9-10x
C12.3(5x-6)-2(2x-5)=11x-10
C13.10x+5(3x-2)=25-10x
C14.6(2x-3)+3(3-5x)=8x-9
C15.3(4x-2)+2(6-2x)=10-6x
C16.5(3-6x)-4(2-2x)=4x-9
B2:tìm cặp số nguyên x, y thỏa mãn
X y+2x+y=0
nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên
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Bài 1:
16:
=>15-30x-8+8x=4x-9
=>-22x+7=4x-9
=>-26x=-16
=>x=8/13
15: \(\Leftrightarrow12x-6+12-4x=10-6x\)
=>8x+6=10-6x
=>14x=4
=>x=2/7
14: \(\Leftrightarrow12x-18+9-15x=8x-9\)
=>-3x-9=8x-9
=>x=0
13: \(\Leftrightarrow10x+15x-10=25-10x\)
=>25x-10=25-10x
=>35x=35
=>x=1
12: \(\Leftrightarrow15x-18-4x+10=11x-10\)
=>11x-8=11x-10(loại)
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Cho: M(x)= -5x^3 - 4x + 3x^4 + 6 + 3x^3 + 5
N(x)= 3x^4 + 4x - x^5 +2x^3 - 5 + x^5
a) thu gọn và sắp xếp theo luỹ thừa giảm
b) P(x)=M(x)+N(x)
Q(x)=M(x)-N(x)
c) chứng tỏ đa thức P(x) ko có nghiệm
Tìm x:
1/3x(\(-\dfrac{4}{3}\)x+1)-4x(x-2)=10
2/5(x2-3x+1)+x(1-5x)=x-2
3/12x2-4x(3x-5)=10x-17
4/4x2-2x+3-4x(x-5)=7x-3
5/-3(x-5)+5(x-1)+3x2=4-x
1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)
=>-8x^2+11x-10=0
=>\(x\in\varnothing\)
2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
=>-14x+5=x-2
=>-15x=-7
=>x=7/15
3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)
=>10x=-17
=>x=-17/10
4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)
=>18x+3=7x-3
=>11x=-6
=>x=-6/11
5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)
\(\Leftrightarrow3x^2+2x+10-4+x=0\)
=>3x^2+3x+6=0
hay \(x\in\varnothing\)
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c) (7x + x – 28) :100 – 38 9.d) 936 – (4x + 5x ) 900e) 280 + 78 : (135 – x :7) 286f) x – 18 :3 16 + 16 : 23. g) (3x – 10) : 10 20.h) (3x – 24 ).73 2.74i) 123 – 5.( x + 4) 38. k) [(6x – 72) :2 – 84].28 5628.m) 2x – 138 23.32.n) [(10 - x).2 + 7]:3 -2 3.Trình bày kết quả rõ ràng nha cảm ơn mọi người rất nhiều❤❤❤
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c) (7x + x – 28) :100 – 38 = 9.
d) 936 – (4x + 5x ) = 900
e) 280 + 78 : (135 – x :7) =286
f) x – 18 :3 = 16 + 16 : 23.
g) (3x – 10) : 10 = 20.
h) (3x – 24 ).73= 2.74
i) 123 – 5.( x + 4) = 38.
k) [(6x – 72) :2 – 84].28 = 5628.
m) 2x – 138 = 23.32.
n) [(10 - x).2 + 7]:3 -2 = 3.
Trình bày kết quả rõ ràng nha cảm ơn mọi người rất nhiều❤❤❤
c) (7x + x – 28) :100 – 38 = 9
<=> 8x - 28 = 4700
<=> x = 591
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d) 936 – (4x + 5x ) = 900
<=> 9x = 36
<=> x = 4
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: Tìm x, biết. a) 7(x – 5) + 2 = 51 k) 2412 : (3x + 147) = |-38| + (-26) b) (43 – 11x).53 = 4.54 l) 4824 : (4x + 137) = |-59| + (-35) c) |-123| - 5(x – 3) = (-28) + 66 m) 7x-4 .6 = 2058 d) 42 – 3(5x + 1) = 35 : 33 n) 27 - |x| = 2.(52 – 24 ) e) |x| - 15 = - 5 o) 3.2x + 2x+3 = 44 f) 2x – 2828 : 14 = 308 p) 95 – 5(x + 3) = 75 : 73 + 21 g) 3x + 3x+1 + 3x+2 = 1053 q) 1300 : [110 – (x – 7)] = 26 h) (11x – 23 ).93 = 4.94 r) 5.(12 – 3x) – 20 = 10
a) 7(x - 5) + 2 = 51
\(\Leftrightarrow\) 7(x - 5) = 51 - 2
\(\Leftrightarrow\) 7(x - 5) = 49
\(\Leftrightarrow\) x - 5 = 49 : 7
\(\Leftrightarrow\) x - 5 = 7
\(\Leftrightarrow\) x = 7 + 5
\(\Leftrightarrow\) x = 12.
Vậy x = 12.
k) 2412 : (3x + 147) = |-38| + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 12
\(\Leftrightarrow\) 3x + 147 = 2412 : 12
\(\Leftrightarrow\) 3x + 147 = 201
\(\Leftrightarrow\) 3x = 201 - 147
\(\Leftrightarrow\) 3x = 54
\(\Leftrightarrow\) x = 54 : 3
\(\Leftrightarrow\) x = 18.
Vậy x = 18.
I) 4824 : (4x + 137) = |-59| + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 24
\(\Leftrightarrow\) 4x + 137 = 4824 : 24
\(\Leftrightarrow\) 4x + 137 = 201
\(\Leftrightarrow\) 4x = 201 - 137
\(\Leftrightarrow\) 4x = 64
\(\Leftrightarrow\) x = 64 : 4
\(\Leftrightarrow\) x = 16.
Vậy x = 16.
c) |-123| - 5(x - 3) = (-28) + 66
\(\Leftrightarrow\) 123 - 5(x - 3) = 38
\(\Leftrightarrow\) 5(x - 3) = 123 - 38
\(\Leftrightarrow\) 5(x - 3) = 85
\(\Leftrightarrow\) x - 3 = 85 : 5
\(\Leftrightarrow\) x - 3 = 17
\(\Leftrightarrow\) x = 17 + 3
\(\Leftrightarrow\) x = 20.
Vậy x = 20.
m) 7x - 4 . 6 = 2058
\(\Leftrightarrow\) 7x - 24 = 2058
\(\Leftrightarrow\) 7x = 2058 + 24
\(\Leftrightarrow\) 7x = 2082
\(\Leftrightarrow\) x = 2082 : 7
\(\Leftrightarrow\) x = \(\frac{2058}{7}\).
Vậy x = \(\frac{2058}{7}\).
Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.
Chúc bạn học tốt !!!
---------------NHANH NHA MK ĐANG CẦN GẤP--------------
-----------------------------------------------------
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
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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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I: thu gọn
M= (-x-8)-(-3x+10)-(x-10)
N=-(x-100)+(-3x+10)-(-x-100)
Q=100-(-4x+1)-(99+x)-(x-1)
Help me
\(M=\left(-x-8\right)-\left(-3x+10\right)-\left(x-10\right)\\ =-x-8+3x-10-x+10\\ =\left(-x+3x-x\right)+\left(-8-10+10\right)\\ =x-8\)
\(N=-\left(x-100\right)+\left(-3x+10\right)-\left(-x-100\right)\\ =-x+100+-3x+10+x+100\\ =\left(-x+-3x+x\right)+\left(100+10+100\right)\\ =-3x+210\\ =3\left(-x+70\right)\)
\(Q=100-\left(-4x+1\right)-\left(99+x\right)-\left(x-1\right)\\ =100+4x-1-99-x-x+1\\ =\left(4x-x-x\right)+\left(100-1-99+1\right)\\ =2x+1\)
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Rút gọn biểu thức:
a) (x 2 – 2x + 2)(x 2 – 2)(x 2 + 2x + 2)(x 2 + 2)
b) (x + 1)2 – (x – 1)2 + 3x 2 – 3x(x + 1)(x – 1)
c) (2x + 1)2 + 2(4x 2 – 1) + (2x – 1)2
d) (m + n)2 – (m – n)2 + (m – n)(m + n)
e) (3x + 1)2 – 2(3x + 1)(3x + 5) + (3x + 5)2
a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)\)
\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)
\(=x^8-16\)
b: Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)
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