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a) A = 3 ( x − y ) 2 − 2 ( x + y ) 2 − ( x − y ) ( x + y ) 2 A = [ ( x − y ) − ( x + y ) ] 2 + 5 ( x − y ) 2 − 5 ( x + y ) 2 2 A = 4 y 2 + 5 [ ( x − y ) − ( x + y ) ] [ ( x − y ) + ( x + y ) ] 2 A = 4 y 2 + 5 [ − 2 y ] [ 2 x ] = 4 y 2 − 20 x y = 4 y ( y − 5 x ) A = 2 y ( y − 5 x )
a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)
\(x-\dfrac{3}{4}=\dfrac{9}{4}\)
=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)
b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)
\(\dfrac{7}{8}:x=\dfrac{5}{2}\)
=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)
c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)
e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)
f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)
\(\dfrac{17}{3}:x=\dfrac{5}{3}\)
=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)
a: =>x-3/4=18/8=9/4
=>x=9/4+3/4=12/4=3
b: =>7/8:x=5/2
=>x=7/8:5/2=7/8*2/5=14/40=7/20
c: x+1/2*1/3=3/4
=>x+1/6=3/4
=>x=3/4-1/6=9/12-2/12=7/12
d: =>12/10-x=2/3
=>6/5-x=2/3
=>x=6/5-2/3=18/15-10/15=8/15
e: =>x*10/3=10/3:17/4=10/3*4/17
=>x=4/17
f: =>17/3:x=13/3-5/2=26/6-15/6=11/6
=>x=17/3:11/6=17/3*6/11=34/11
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
a) A = (x+3)2 + (x-3)(x+3) - 2(x+2)(x - 4)
= (x + 3)(x + 3) + (x - 3)(x + 3) - 2[x(x - 4) + 2(x - 4)]
= x(x + 3) + 3(x + 3) + x(x + 3) - 3(x + 3) - 2[x2 - 4x + 2x - 8]
= x2 + 3x + 3x + 9 + x2 + 3x - 3x - 9 - 2(x2 - 2x - 8)
= x2 + 3x + 3x + 9 +x2 + 3x - 3x - 9 - 2x2 + 4x + 16
= (x2 + x2 - 2x2) + (3x + 3x + 3x - 3x + 4x) + (9 - 9 + 16) = 10x + 16
Thay x = -1/2 vào biểu thức trên ta có : \(10\cdot\left(-\frac{1}{2}\right)+16=-5+16=11\)
b) \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)
\(B=9x^2+24x+16-x\left(x+4\right)+4\left(x+4\right)-10x\)
\(B=9x^2+24x+16-x^2-4x+4x+16-10x\)
\(B=\left(9x^2-x^2\right)+\left(24x-4x+4x-10x\right)+\left(16+16\right)\)
\(B=8x^2+14x+32\)
Thay x = -1/10 vào biểu thức trên ta có : \(B=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)
c) \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)
\(C=x^2+2x+1-\left(2x-1\right)\left(2x-1\right)+3\left(x^2-4\right)\)
\(C=x^2+2x+1-2x\left(2x-1\right)+1\left(2x-1\right)+3x^2-12\)
\(C=x^2+2x+1-4x^2+2x+2x-1+3x^2-12\)
\(C=\left(x^2-4x^2+3x^2\right)+\left(2x+2x+2x\right)+\left(1-1-12\right)\)
\(C=6x-12\)
Thay x = 1 vào biểu thức ta có : C = 6.1 - 12 = 6 -12 = -6
Còn bài kia làm nốt đi
1. Tính:
a)(x+4) (x^2 - 4x+16)=(x+4)(x2-4x+42)=x3+64
b)(x-3y) (x^2 + 3xy+3y^2)=x3-(3y)3
c)(x-3)^2 -(x+3)^3=(x2-6x+9)-(x3+9x2+27x+27)=-8x2-33x-18-x3
d)(x^2-1)^3 - (x^2 +1)^3=(x2-1-x2-1)[(x2-1)2+(x2-1)(x2+1)+(x2+1)2]
=-2(x4-2x2+1+x4-1+x4+2x2+1)
=-2(3x4+2)=-6x4-4
a/ \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)
\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8-x^3-2x=0\)
\(\Leftrightarrow-2x=-8\)
\(\Leftrightarrow x=4\)
Vậy .....................
b/ \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)
\(\Leftrightarrow-3x^4+3x^2=0\)
\(\Leftrightarrow3x^2\left(-x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2=0\\-x^2+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x^2=-1\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy pt có 3 nghiệm là \(\left\{{}\begin{matrix}x=-1\\x=0\\x=1\end{matrix}\right.\)
| x | - 1 = 2
| x | = 2 + 1
|x| = 3
X = 3 hoặc x = -3
chọn D. x = 3 hoặc x = - 3
Đáp án đúng : D