(2x-5)(3x+7)=4x^2-25
Giải các phương trình sau:
a/ 3x – 2 = 2x – 3
b/ 7 – 2x = 22 – 3x
c) 8x – 3 = 5x + 12
d/ x – 12 + 4x = 25 + 2x – 1
e/ x + 2x + 3x – 19 = 3x + 5
a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)
Vậy: \(S=\left\{-1\right\}\)
b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)
Vậy: \(S=\left\{15\right\}\)
c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)
Vậy: \(S=\left\{5\right\}\)
d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)
Vậy: \(S=\left\{12\right\}\)
e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=12
=>x=4
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=36
=>x=12
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
1)6x-8=3x+1
2)12-10x=25-30x
3)3(2x+3)-2(4x-5)=10x+21
4)5(5x-3)-3(2x-4)11-5x
5)4(2-3x)-5(1-2x)=4-6x
6)8(4x-3)-3(2-3x)=13-40x
7)10x-5(1-4x)=5x-11
8)-3(3-4x)-5(4-3x)=12x-50
9)-2(20x-3)-3(4x-5)=9-2(2x-3)
10)-5(2-3x)+3(5-2x)=3x+3(3-5x)
1)6x-8=3x+1
6x-3x=1+8
3x=9
x=3
Vậy x=3
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
Giải pt sau:
a, 3 - 4x( 25 - 2x ) - 8x2 + x - 300
b, 2( 1 -3x )/5 - 2+ 3x/10 = 7- 3( x + 1)/4
c, 5x + 2 /6 - 8x - 1/3 = 4x + 2/5 - 5
d, 3x + 2/3 - 3x + 1/6 = 2x + 5/3
Help me
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
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
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(m,x^2-11x+28=0\)
\(\Leftrightarrow x^2-4x-7x+28=0\)
\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)
\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)
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 \(x=2\)
#\(Toru\)
Bài 1:
a)-2x-3.( x-17 )= 34-2.( -x+25 )
b)17x+3.( -16x-37 )= 2x+43-4x
c){-3x+2. [ 45-x-3 ( 3x+7 ) -2x]+4x} = 55
g)-103-57: [ 5.( 2x-1 )2-( -9 )0 ]
Bài 2:
a)-5.( -x+7 )-3.( -x-5 )=-4.( 12-x )+48
b)7.( -x-7 )-4.( -2x-11 )=7.( 4x+10 )+9
c)-2.( 15-3x )-4.( -7x+8 )=-5-9.( -2x+1 )
d)5.( -3x-7 )-4.( -2x-11 )=7.( 4x+10 )+9
e)4( x-2 )2-13=( -3 )2.2-11.( -3 )
f)-52-( 2x-1 )3=( -13 ).( -3 )
Các bạn giúp mình nhé càng nhanh càng tốt nhà
(5x-1). (2x+3)-3. (3x-1)=0
x^3 (2x-3)-x^2 (4x^2-6x+2)=0
x (x-1)-x^2+2x=5
(3x+2)(x-1)-3 (5x+2)+5 (11-4x)=25
8 (x-2)-2 (3x-4)=25
(3x+4). (5x-1)+(5x+2). (1-3x)+2=0
(5x-1). (2x+7)-(2x-3). (5x+9)
4 (x-1). (X+5)-(x+5). (X+2)=3. (X-1)(x+2)
2x^2+3 (x-1). (X+1)=5x(x+1)
4. (18-5x)-12 (3x-7)=1825. (2x-16)-6 .(x+4)
1/2x. (2/5-4x)+(2x+5).x=-13/2
Nhiều các bạn giả đùm mình nha
Thanh nhiều
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
\(\left(2x-5\right).\left(3x+7\right)=4x^2-25\)
Ta có : ( 2x - 5).(3x+7 ) = 4x2 -25
<=>6x2 + 14x - 15x -35 = 4x2 -25
<=> 2x2 -x -
Mình xin lỗi bạn nha ,do bất cẩn quá nên mình lỡ nhấn gửi câu trả lời ,để mình làm lại nhé !!!
Ta có :( 2x -5 )(3x + 7)=4x2 -25
<=> 6x2 + 14 - 15x-35 =4x2 -25
<=> 2x2 -x - 10 = 0
<=> ( 2x-5)(x+2)=0
<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-2\end{cases}}\)
Vậy x = \(\left\{\frac{5}{2};-2\right\}\)
\(\left(2x-5\right)\left(3x+7\right)=4x^2-25\)
\(6x^2+14x-15x-35=4x^2-25\)
\(6x^2-x-35=4x^2-25\)
\(6x^2-x-35-4x^2+25=0\)
\(2x^2-x-10=0\)
\(2x^2+4x-5x-10=0\)
\(2x\left(x+2\right)-5\left(x+2\right)=0\)
\(\left(2x-5\right)\left(x+2\right)=0\)
\(\orbr{\begin{cases}2x-5=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=5\\x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-2\end{cases}}}\)
1. a. 4x-20=0
b. 2x+x+12=0
c. x-5=3-x
d. 7-3x=9-x
2.
a. 7+2x=22-3x
b. 8x-3=5x+12
c. x-12+4x=25+2x-1
d. x+2x+3x-19=3x+5
e. 7-(2x+4)=-(x+4)
f. (x-1)-(2x-1)=9-x