Rút gọn biểu thức
(a-b+c-d)-(a+b+c+d)
(-a+b-c)+(a-b)-(a-b+c)
-(a-b-c)+(b-c+d)-(a+b+d)
Tìm số nguyên x,biết
x+x+x+82=-2-x
5 nhan (-40) nhan x=-100
(-1) nhan (-3) nhan (-6) nhan x=36
gia tri tuyet doi cua
1-4x=7
x nhan(x-2)=0
x nhan(x-2) lon hon 0
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\(\left(a-b+c\right)-\left(a+c\right)\)
\(=a-b+c-a-c\)
\(=\left(a-a\right)+\left(c-c\right)-b\)
\(=0+0-b\)
\(=0-b\)
\(=-b\)
1) (a - b + c) - (a + c)
= a - b + c - a - c
= (a - a) - b + (c - c)
= 0 - b + 0 = -b
2) (a + b) - (b - a) + c
= a + b - b + a + c
= (a + a) + (b - b) + c
= 2a + 0 + c = 2a + c
3) -(a + b - c) + (a - b - c)
= -a - b + c + a - b - c
= (-a + a) - (b + b) + (c - c)
= 0 - 2b + 0 = -2b
4) a(b + c) - a(b + d)
= ab + ac - ab - ad
= (ab - ab) + a(c - d)
= 0 + a(c - d) = a(c - d)
5) tự lm
a: \(=5x^4-4x^3y+10x^3y-8x^2y^2-25x^2y^2+20xy^3-15xy^3+12y^4\)
\(=5x^4+6x^3y-33x^2y^2+5xy^3+12y^4\)
b: \(=\left(x^2-x-2\right)\left(2x-1\right)\)
\(=2x^3-x^2-2x^2+x-4x+2\)
\(=2x^3-3x^2-3x+2\)
c: \(=8x^3+y^3\)
d: \(=a^4-b^4\)
Bài 1 câu g bạn kia làm sai mình sửa lại nhá
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
Để mình làm tiếp cho :))
Bài 2 :
Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)
\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)
\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)
\(=37,5.10-7,5.10\)
\(=10.30=300\)
Câu b : \(35^2+40^2-25^2+80.35\)
\(=\left(35^2+80.35+40^2\right)-25^2\)
\(=\left(30+45\right)^2-25^2\)
\(=75^2-25^2\)
\(=\left(75+25\right)\left(75-25\right)\)
\(=100.50=5000\)
Bài 3 :
Câu a : \(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)
Câu b : \(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)
Câu c :
\(x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2\left(x-3\right)+27-9x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)
Bài 4 :
Câu a :
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
Câu b :
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Câu c :
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
Câu d :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
x+x+x+82=-2-x
=>x+x+x+x=-2-82
4x=-84
x-84/4
x=-21
(a-b+c-d)-9a+b+c+d)=a-b+c-d-a-b-c-d
=(a-a)-(b+b)+(c-c)-(d-d)
=0-2b+0-2d
=-2b-2d
(-a+b-c)+(a-b)-(a-b+c)
=-a+b-c+a-b-a+b-c
=(-a+a-a)+(b-b+b)-(c+c)
=-a+b-2c
-(a-b-c)+(b-c+d)-(a+b+d)
=-a+b+c+c-c+d-a-b-d
=(-a-a)+(b-b)+(c+c-c)+(d-d)
=-2a+0+c+0
=-2a+c
x+x+x+82=-2-x
3x+82=-2-x
3x+x=-2-82
4x=-84
x=-84:4=-21
5.(-40).x=-100
-200.x=-100
x=-100:(-200)
x=0,5
-1.(-3).(-6).x=36
-18.x=36
x=36:(-18)
x=-2
|1-4x|=7
Xảy ra hai trường hợp:1-4x=7=>4x=1-7=-6=>x=-6:4=-1,5
1-4x=-7=>4x=1+7=8=>x=8:4=2
Vì x là số nguyên nên ta chỉ có một đáp án đó là:2
x.(x-2)=0+> x=0 hoặc x=2
x.(x-2)>0=>x=-1;-2;-3;-4;...