rut gon bieu thuc
(x+1)3+(x-1)3+x3-3x(x+1)(x-1)
Những câu hỏi liên quan
cho bieu thuc A = /x-1/ + 3x-7
a. rut gon bieu thuc
b. tinh A khi x=3;x=-5
Cho bieu thuc: ( x-1/ x+1 - x-1/x+1) : 2x / 3x - 3
a, Tim dieu kien xac dinh cua bieu thuc P
b, Rut gon bieu thuc P
c, Tim x thuoc z de P nhan gia tri nguyen.
Đề bài sai rồi bạn ! Mình sửa :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)
\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-6}{x+1}\)
c) Để P nhận giá trị nguyên
\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)
\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
Ta loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
rut gon cac bieu thuc
2x(2x+1)2 - 3x(x+3)(x-3) - 4x (x+1) 2
Ta có 2x(2x + 1)2 - 3x(x + 3)(x - 3) - 4x(x + 1)2
= 2x(4x2 + 4x + 1) - 3x(x2 - 9) - 4x(x2 + 2x + 1)
= 8x3 + 8x2 + 2x - 3x3 + 27x - 4x3 - 8x2 - 4x
= 8x3 - 3x3 - 4x3 + 8x2 - 8x2 + 2x + 27x - 4x
= x3 + 25x
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a oi hinh nhu sai r con +16x2 nua co a , anh tinh lai ho e duoc kh
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Kiều Trinh Vũ ko có + 16x2 nhá vì 8x2 - 8x2 = 0 nhá
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Xem thêm câu trả lời
B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy
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1) Rut gon cac bieu thuc sau :
a) (x-3)(x2+3x+ 9)-(54+x3)
b) (3x+y)(9x2-3xy +y2)-(3x-y)(9x2+3xy+y2)
2, Dien cac don thuc thich hop vao cho trong
a, (x+3y) (... - ... + ...) = x^3 +27y^3
b, (2x- ....) (... + 6xy + ... +...) = 8x^3 - 27y^3
\(1.\)
\(a.\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\)
\(=x^3-27-54-x^3\)
\(=-81\)
\(b.\)
\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)\)
\(=27x^3+y^3-27x^3+y^3\)
\(=2y^3\)
\(2.\)
\(a.\)
\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
\(b.\)
\(\left(2x-3y\right)\left(4x^2+6xy+9y^3\right)=8x^3-27y^3\)
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1) a) \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)
\(=\left(x^3-3^3\right)-\left(54+x^3\right)\\ =\left(x^3-27\right)-54-x^3\\ =-27-54\\ =-81\)
b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left[\left(3x\right)^3+y^3\right]-\left[\left(3x\right)^3-y^3\right]\\ =2y^3\)
2) a) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)
b) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)
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rut gon bieu thuc tren (x-1)^3-(x-1).(x^2+x+1)
Lời giải:
$(x-1)^3-(x-1)(x^2+x+1)=(x-1)[(x-1)^2-(x^2+x+1)]=(x-1)(x^2-2x+1-x^2-x-1)=(x-1)(-3x)=-3x(x-1)$
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Rut gon bieu thuc sau: 2x*(2x-1)2-3x*(x+3)2-4x*(x+1)2
rut gon bieu thuc a,(3x+2)^2+4x-3^2+2(5x-2)(5x+2)-75x^2
b,(x-2)^3+(2x+1)^3++2(x+2)(1-x)-9x^3+2x
a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)
\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)
\(=-19x^2+16x-4\)
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rut gon bieu thuc
A=3 I x-1 I -2 I 5-3x I
B=4 I x-3 I +2 I 2x -1 I +I 4-3x I