Tìm x
A)2x-3/4=-5/8
B)3x-1/7=1:14-1/21
C)-2/5.x+1/4=1+1/3
D)1:6.2x-3=1/2.(-x+1/4)-2/3
Tìm x
a) -x/27 -1 = 2/3
b) x - 4 = -14/35 : 20/-21
c) x + 2/3 = -1/12 . -4/5
d) 1/5 - 3/5x = -15/14 :20/-21
e) -3/7x = 3/5 . 28/9
f)1/2x + 3/5x = -2/3
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
Bài1:Rút gọn
a,(4x-5)(3x+2)-(7-3x)(x+2)
b,(-2x+1)(x-5)-3(x-2)(x+1)
c,(x^2-7)(x-5)+(3x^2+5)(2x-4)
d,(x^2+3x-2)(x+4)-4x(x-5)
Bài2:Tìm xbiết
a,(x-4)(x+3)-(x+1)(x-5)=8
b,(3x-2)(x+1)-3x(x+7)=13
c,(x+5)(x-5)-x(x+2)=9
d,(x-1)(x^2+x+1)-x(x^2-3)=1
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
Tìm x:
a) 2/5 = 3/4 : x = -1/2
b) 5/7 - 2/3 . x = 4/5
c) 1/2x + 3/5x = -2/3
d) 4/7x - x = -9/14
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
bài 1: tìm x
a) 8/23 . 46/24 = 1/3 . x
b) 1/5 : x = 1/5 - 1/7
c) 4/9 - (x - 1/2) mũ 2 = 1/3
d)3,2 . x - (4/5 + 2/3) : 3 2/3 = 7/10
` 8/23 . 46/24 =1/3 .x`
`=>8/23 . 23/12 =1/3 . x`
`=> 1/3 . x=2/3`
`=>x=2/3 : 1/3`
`=>x=2/3 . 3`
`=> x= 6/3`
`=>x=2`
`----`
`1/5 : x= 1/5-1/7`
`=>1/5 : x= 7/35 - 5/35`
`=> 1/5 :x= 2/35`
`=>x= 1/5 : 2/35`
`=>x=1/5 . 35/2`
`=>x=7/2`
`----`
`4/9 - (x-1/2)^2 =1/3`
`=> (x-1/2)^2 =4/9-1/3`
`=> (x-1/2)^2 =4/9- 3/9`
`=> (x-1/2)^2 =1/9`
`=> (x-1/2)^2 = (+- 1/3)^2`
`@ TH1`
`x-1/2=1/3`
`=>x=1/3+1/2`
`=>x= 2/6 + 3/6`
``=>x= 5/6`
`@ TH2`
`x-1/2=-1/3`
`=>x=-1/3 +1/2`
`=>x= -2/6 + 3/6`
`=>x=1/6`
`----`
`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`
`=> 3,2 . x-22/15 : 11/3 = 7/10`
`=> 3,2 . x-22/15 = 7/10 . 11/3`
`=> 3,2 . x-22/15 =77/30`
`=> 3,2 .x= 77/30 + 22/15`
`=> 3,2 .x=121/30`
`=>x= 121/30. 5/16`
`=>x= 121/96`
Giải các bất phương trình sau:
a) 2(3x + 1) - 4(5 - 2x) > 2(4x - 3) - 6
b) 9x2 - 3(10x - 1) < (3x - 5)2 - 21
c) \(\dfrac{x-1}{2}+\dfrac{x-2}{3}+\dfrac{x-3}{4}>\dfrac{x-4}{5}+\dfrac{x-5}{6}\)
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
Bài 1 : giải phương trình
a) (x-2)(x+2)-(2x+1)2=x(2-3x)
b) 2x(x+2)2-8x2=2(x-2)(x2+2x+4)
c) (x-2)3+(3x-1)(3x+1)=(x+1)3
d) 5(2x-3)-4(5x-7)=19-2(x+1)2
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
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
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
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
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 \(x=2\)
#\(Toru\)
Tìm x :( bài 14 trang 11 sách bồi dưỡng năng lực tự học toán 8)
Câu 2 : (2x+3)2+(x-1)*(x+1)=5*(x+2)2-(x-5)*(x+1)+(x+4)2
Câu 3 : (-x+5)*(x-2)+(x-7)*(x+7)=(3x+1)2-(C)*(3x+2)
Câu 4 : (5x-1)*(x+1)-2(x-3)2=(x+2)*(3x-1)-(x+4)2+(x2-x)
Câu 5 : (4x-1)2-(3x+2)*(3x-2)=(7x-1)*(x+2)+(2x+1)2-(3x+2)
Câu 6 : (2x+3)2-(5x-4)*(5x+4)=(x+5)2-(3x-1)*(7x+2)-(x2-1+1)
Câu 7 : (1-3x)2-(x-2)*(9x+1)=(3x-4)*(3x+4)-9(x+3)2
Câu 8 : (3x+4)*(3x-4)-(2x+5)=(x-5)+(2x+1)2-(x2-2x)+(x-1)2
Câu 9 : (x-7)*(x+1)-(x-3)2=(3x-5)*(3x+5)-(3x+1)+(x-2)2-x2
Câu 10 : -5(x+3)2+(x-1)*(x+1)+(2x-3)=(5x-2)2-5x(5x+3)