a,|x+1|+|x-3|=5
b,|2x+3|+|x+7|=1
c,|4x-5|+|2x+3|=3
d,|x-1/2|+|3x-2|=1
Cứu mình
a) x + 7 ⋮ x + 2
b) 2x + 5 ⋮ x + 1
c) 3x - 2 ⋮ x + 3
d) 12x + 1 ⋮ 3x + 2
e) x2 + 3x + 5 ⋮ x + 3
f) X2 - 2x + 3 ⋮ x + 2
a: \(x+7⋮x+2\)
=>\(x+2+5⋮x+2\)
=>\(5⋮x+2\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
b: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
c: \(3x-2⋮x+3\)
=>\(3x+9-11⋮x+3\)
=>\(-11⋮x+3\)
=>\(x+3\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-2;-4;8;-14\right\}\)
d: \(12x+1⋮3x+2\)
=>\(12x+8-7⋮3x+2\)
=>\(-7⋮3x+2\)
=>\(3x+2\in\left\{1;-1;7;-7\right\}\)
=>\(3x\in\left\{-1;-3;5;-9\right\}\)
=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)
e: \(x^2+3x+5⋮x+3\)
=>\(x\left(x+3\right)+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
f: \(x^2-2x+3⋮x+2\)
=>\(x^2+2x-4x-8+11⋮x+2\)
=>\(11⋮x+2\)
=>\(x+2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{-1;-3;9;-13\right\}\)
a,|7 - 2x| + 7 = 2x
b,| 1 - x | = 4x + 1
c, | x - 1/3 | + 4/5 = | ( 3,2) + 2/5 |
d,| x - 7 | + 2x + 5 = 6
e, 3x - | 2x - 1 | = 2
a: \(\left|7-2x\right|+7=2x\)
=>\(\left|2x-7\right|+7=2x\)
=>\(\left|2x-7\right|=2x-7\)
=>2x-7>=0
=>\(x>=\dfrac{7}{2}\)
b: \(\left|1-x\right|=4x+1\)
=>\(\left|x-1\right|=4x+1\)
=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)
d: \(\left|x-7\right|+2x+5=6\)
=>\(\left|x-7\right|=6-2x-5=-2x+1\)
=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
e: 3x-|2x-1|=2
=>|2x-1|=3x-2
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
giải phương trình:
a)(2x-3)(2x+3)=4x(x-5)-3x
b)(2x+1)(4x-3)=4x^2-1
c)3x/x-2+x/5-x-2x^2+5/x^2-7x+10=0
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
a/ x+2.y-1=5
b/ 2x+1.y-3=-7]
c x.y-x+y-1=3
d/ x.y+3x+y=2
Đề yêu cầu làm gì, em ghi rõ ra nhé!
b: (2x+1)(y-3)=-7
=>(2x+1;y-3) thuộc {(1;-7); (-7;1); (-1;7); (7;-1)}
=>(x,y) thuộc {(0;-4); (-4;4); (-1;10); (3;1)}
c: xy-x+y-1=3
=>(y-1)(x+1)=3
=>(x+1;y-1) thuộc {(1;3); (3;1); (-1;-3); (-3;-1)}
=>(x,y) thuộc {(0;4); (2;2); (-2;-2); (-4;0)}
d: =>x(y+3)+y+3=5
=>(x+1)(y+3)=5
=>(x+1;y+3) thuộc {(1;5); (5;1); (-1;-5); (-5;-1)}
=>(x,y) thuộc {(0;2); (4;-2); (-2;-8); (-6;-4)}
Bài 1.khai triển HĐT
a,(3x-4)^2 b,(1+4x)^2 c,(2x+3)^3
d,(5-2x)^3 e,49x^2-25 f,1/25-81y^2
Bài 2.Tìm x biết:Viết đầy đủ
a,(x-5)^2-(x+7)(x-7)=8 b,(2x+5)^2-4(x+1)(x-1)=10
Bài 3.Tìm GTLN,GTNN của các biểu thức sau
a,A=x^2-6x+19 b,B=-x^2+8x-20
c,C=4x^2+12x+100 d,D=25+4x-x^2
Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
Bài 3.
\(a,A=x^2-6x+19\)
\(=x^2-6x+9+10\)
\(=\left(x^2-2\cdot x\cdot3+3^2\right)+10\)
\(=\left(x-3\right)^2+10\)
Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+10\ge10\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: \(Min_A=10\) khi \(x=3\)
\(b,B=-x^2+8x-20\)
\(=-x^2+8x-16-4\)
\(=-\left(x^2-8x+16\right)-4\)
\(=-\left(x^2-2\cdot x\cdot4+4^2\right)-4\)
\(=-\left(x-4\right)^2-4\)
Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-4\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-4\right)^2-4\le-4\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy \(Max_B=-4\) khi \(x=4\)
\(c,C=4x^2+12x+100\)
\(=4x^2+12x+9+91\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot3+3^2\right]+91\)
\(=\left(2x+3\right)^2+91\)
Ta thấy: \(\left(2x+3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+3\right)^2+91\ge91\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(Min_C=91\) khi \(x=\dfrac{-3}{2}\)
\(d,D=25+4x-x^2\)
\(=-x^2+4x-4+29\)
\(=-\left(x^2-2\cdot x\cdot2+2^2\right)+29\)
\(=-\left(x-2\right)^2+29\)
Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+29\le29\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy \(Max_D=29\) khi \(x=2\)
#\(Toru\)
1. Thu gọn biểu thức
a) (x-3) ² + 3x (x-5)
b) (3x+2) ² - (x+3) (x-3)
2. Tìm x biết a) (x+4) ² - (x+2) (x-2)=5
b) (3x-1) ² _ (2x-3) (4x+1)= 5+x ²
1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
bài 1
a)(x-1)(x+2)-(x-3)(x+1)=5x-3
b)(2x-1)(x+3)-(x-2)(x+2)=3x+1
c)x^2(x-1)-x(x-1)(x+1)=0
d)4x(x-5)-(2x-3)(2x+3)=9
Lời giải:
a.
a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$
$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$
$\Leftrightarrow 3x+1=5x-3$
$\Leftrightarrow 4=2x$
$\Leftrightarrow x=2$
b.
$(2x-1)(x+3)-(x-2)(x+3)=3x+1$
$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$
$\Leftrightarrow x^2+5x+1=3x+1$
$\Leftrightarrow x^2+2x=0$
$\Leftrightarrow x(x+2)=0$
$\Leftrightarrow x=0$ hoặc $x=-2$
c.
$x^2(x-1)-x(x-1)(x+1)=0$
$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$
$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$
$\Leftrightarrow (x-1)(-x)=0$
$\Leftrightarrow x-1=0$ hoặc $-x=0$
$\Leftrightarrow x=1$ hoặc $x=0$
d.
$4x(x-5)-(2x-3)(2x+3)=9$
$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$
$\Leftrightarrow -20x=0$
$\Leftrightarrow x=0$
a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow2x=4\)
hay x=2
b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)
\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)
\(\Leftrightarrow4x^2-20x-4x^2+9=9\)
hay x=0
giải các phương trình sau :
a, x^2 - 10x = -25
b, 4x^2 - 4x = -1
c, ( 1 - 2x )^2 = ( 3x - 2 )^2
d, ( x - 2 )^3 + ( 5 - 2x )^3 = 0
\(a,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\\ b,\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow x=1\\ c,\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\\ \Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\\ \Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2\right)^3=-\left(5-2x\right)^3\\ \Leftrightarrow x-2=-\left(5-2x\right)=2x-5\\ \Leftrightarrow x=3\)