Bài 3 : Tìm x Z biết.
a) x(x + 2) = 0 e) 7x – 13 = 3 2 .4
b) 5 – 2x = -7 f) 155 – 5(x + 3) = 80
c) (x + 3)(x – 4) = 0 g) 119 + 3 3 .x = 2 3 . 5 2
d) – 32 – 4(x – 5) = 0 h) 3(2x + 1) – 19 = 14
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
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
a,4x-10=0 b, 7-3x=9-x c, 2x-(3-5x) = 4(x+3)
d, 5-(6-x)=4(3-2x) e, 4(x+3)=-7x+17 f, 5(x-3) - 4=2(x-1)+7
g, 5(x-3)-4=2(x-1)+7 h,4(3x-2)-3(x-4)=7x+20
`a,4x-10=0 `
`<=> 4x=10`
`<=>x=10/4`
`<=>x=5/2`
`b, 7-3x=9-x `
`<=>-3x+x=9-7`
`<=>-2x=2`
`<=>x=-1`
`c, 2x-(3-5x) = 4(x+3)`
`<=>2x-3+5x=4x+12`
`<=>2x+5x-4x=12+3`
`<=>3x=15`
`<=>x=5`
`d, 5-(6-x)=4(3-2x) `
`<=>5-6+x=12-8x`
`<=>x+8x=12-5+6`
`<=>9x=13`
`<=>x=13/9`
`e, 4(x+3)=-7x+17 `
`<=>4x+12=-7x+17`
`<=>4x+7x=17-12`
`<=>11x=5`
`<=>x=5/11`
`f, 5(x-3) - 4=2(x-1)+7`
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`g, 5(x-3)-4=2(x-1)+7 `
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`h,4(3x-2)-3(x-4)=7x+20`
`<=>12x-8-3x+12=7x+20`
`<=>12x-3x-7x=20+8+12`
`<=>2x=40`
`<=>x=20`
Giải các phương trình tích sau: Mng giúp em với ạ.
a) (3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c) 2x(x – 3) + 5(x – 3) = 0 d) (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
e) (x + 2)(3 – 4x) = x2 + 4x + 4 f) x(2x – 7) – 4x + 14 = 0
g) (2x – 5)2 – (x + 2)2 = 0 h) (x2 – 2x + 1) – 4 = 0
i) 3x2 + 2x – 1 = 0 k) x2 – 5x + 6 = 0
l) x2 – 3x + 2 = 0 m) 2x2 – 6x + 1 = -3
a: (3x-2)(4x+5)=0
=>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: (2,3x-6,9)(0,1x+2)=0
=>2,3x-6,9=0 hoặc 0,1x+2=0
=>x=3 hoặc x=-20
c: =>(x-3)(2x+5)=0
=>x-3=0 hoặc 2x+5=0
=>x=3 hoặc x=-5/2
A) 2(x-3)=5(x+4)
B) 8x-3/5=2x+8/8
X-5x+2/6=7-3x/4
C) 7x-3/5=5x+7/7
D) (3x-5)(7x+5)=0
E) (½x -3/4 )(5-2x)=0
G)( 2x+7)(x-5)(5x+1)
H) x+1/x-3 - 1/x-1=2/(x-1)(x-3)
Tìm x biet:
a,5/2x+x=32/15
b,x-7/2x=-32/15
c,-7/3x-7/4x+11/3=18/5
d,-5/6-3x=2/3
e,2x-5/9=-2/3
f,1/2-2/3x=7/12
g,7-|2x-3|=4
h,(19x+2.5^2):14=(13-8)^2-4^2
i,3/4x-2/3=2/7.1/6+5/7.1/6
k,(x+1/2).(2/3-2x)=0
l,x/2013-1/10-1/15-1/21-........-1/120=5/8
c)2x-49=5.3^2
d)2^x=32
e)4^x:16=4
f)(x-1)^3=8
g)7x-x=5^21:5^19+3.2^2-7^0
h)64<x^3<bằng 125
i)25<bằng 5^x<bằng625
c) 2x - 49 = 5 . 32
2x - 49 = 5 . 9
2x - 49 = 45
2x = 45 + 49
2x = 94
x = 94 : 2
x = 47
a) -5 (2x + 6 ) +b | 3x + 9 | = 7x
b) 7(x - 4 ) - | 4x - 12 | = 0
c) | x - 5 | - | x - 3 | =0
d)| x + 1| + | -x + 12 | = 2x -7
e) | x + 2| - | x + 6 | - (-15) = 0
f) | 2x- 10 | + | 3y - 4 | = 0
g) | - 3x + 1 | - | 2x + 8 | + 3x = 0
h) 3( x - 1 ) - 5 | x + 3 | = 0
Mọi người giúp mình nhanh nhé ! Mình đang cần gấp
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^