1.Cho biểu thức M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+\(x^2\)
Tính M theo a,b,c biết rằng \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c\)
2.Cho a+b+c=2p.Cm
\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
M = \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
Tính M theo a, b, c biết rằng x = \(\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c\)
Ta có \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c=\dfrac{a+b+c}{2}\)
Suy ra
M = (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) + x2
= x2 - ax - bx + ab + x2 - bx - cx + bc + x2 - ax - cx + ac + x2
= 4x2 - 2ax - 2bx - 2cx + ab + bc + ac
= (2x)2 - 2x(a + b + c) + ab + bc + ac
= \(\left(2\cdot\dfrac{a+b+c}{2}\right)^2-\left(2\cdot\dfrac{a+b+c}{2}\right)\left(a+b+c\right)+ab+bc+ac\)
= ab + bc + ac
Cho biểu thức
M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x2
Tính M theo a,b,c,biết rằng x=\(\dfrac{1}{2}\) a+\(\dfrac{1}{2}b+\dfrac{1}{2}c\)
(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) + x2
= (x2-ax-bx+ab) + (x2-bx-cx+bc) + (x2-cx-ax+ac) + x2
= 4x2 - 2ax - 2bx + ab + bc + ac
Thay a+b+c = 2x, ta được:
M = 4x2 - 2x(a+b+c) + ab + bc + ac
M = 4x2 - 2x.2x + ab + bc + ac
M = ab + bc + ac
Vậy => đcpcm
Bài 1: Cho biểu thức
B = \(\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
a) Rút gọn B
b) Tìm số nguyên y để \(\dfrac{2B}{2y+3}\) có giá trị nguyên
c) Tìm số nguyên y để B lớn hơn hoặc bằng 1
Bài 2: Cho \(x+\dfrac{1}{x}=3\). Tính giá trị biểu thức
a) A = \(x^2+\dfrac{1}{x^2}\) b) B = \(x^3+\dfrac{1}{x^3}\)
Bài 3: Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=2; \dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\)
Tính giá trị biểu thức D = \(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2\)
Bài 4: Cho a, b, c từng đôi một khác nhau thỏa mãn \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
Tính giá trị biểu thức C = \(\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
Bài 5: Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2; \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
Chứng minh: a + b + c = abc
Bài 1 rút gọn bc tự làm :
\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)
\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)
\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)
\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)
\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)
Bài 2 )
a ) \(x+\dfrac{1}{x}=3\)
\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)
b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
Bài 1: Cho biểu thức
B = \(\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
a) Rút gọn B
b) Tìm số nguyên y để \(\dfrac{2B}{2y+3}\) có giá trị nguyên
c) Tìm số nguyên y để B lớn hơn hoặc bằng 1
Bài 2: Cho \(x+\dfrac{1}{x}=3\). Tính giá trị biểu thức
a) A = \(x^2+\dfrac{1}{x^2}\) b) B = \(x^3+\dfrac{1}{x^3}\)
Bài 3: Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=2; \dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\)
Tính giá trị biểu thức D = \(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2\)
Bài 4: Cho a, b, c từng đôi một khác nhau thỏa mãn \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
Tính giá trị biểu thức C = \(\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ac}+\dfrac{c^2}{c^2+2ab}\)
Bài 5: Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2; \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
Chứng minh: a + b + c = abc
1. Cho \(a^3+b^3+c^3=3abc\) (a+b+c ≠0)
Tính giá trị biểu thức:
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
2. Rút gọn
a) \(\dfrac{x^3+x^2-6x}{x^3-4x}\)
b) \(\dfrac{x^2+8x+7}{x^3+2x^2+x}\)
Bài 1:
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
Bài 2:
a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)
1) Cho \(a+b+c=2p\), Chúng minh hằng đẳng thức
\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
2) Cho biểu thức
\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
Tính M theo a,b,c biết rằng \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\)
HELP ME!!!!!!!!!! NHANH NHANH GIÙM MK NHA
1. Phân tích đa thức thành nhân tử:
\(x^5-x^4+\left(y+2\right)x^3+\left(y-2\right)x^2+yx+y^2\)
2. Cho các số dương thỏa mãn:
\(\dfrac{b+c}{a^2}+\dfrac{c+a}{b^2}+\dfrac{a+b}{c^2}=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Tính giá trị biểu thức sau: \(P=\left(a-b\right)^{2009}+\left(b-c\right)^{2009}+\left(c-a\right)^{2009}\)
3. Cho x,y,x đôi một khác nhau và khác 0. Chứng minh rằng nếu:
\(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\) thì ta có:
\(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
1.
\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)
\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)
\(=\left(x^3-x^2+3x\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)
Hay đa thức trên có thể phân tích thành:
\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)
Dựa vào đó em tự tách cho phù hợp
2.
\(VT=a\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+b\left(\dfrac{1}{a^2}+\dfrac{1}{c^2}\right)+c\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(VT\ge\dfrac{2a}{bc}+\dfrac{2b}{ac}+\dfrac{2c}{ab}=2\dfrac{a^2+b^2+c^2}{abc}\)
\(VP=\dfrac{2\left(ab+bc+ca\right)}{abc}\)
\(\Rightarrow\dfrac{ab+bc+ca}{abc}\ge\dfrac{a^2+b^2+c^2}{abc}\)
\(\Rightarrow ab+bc+ca\ge a^2+b^2+c^2\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\le0\)
\(\Rightarrow a=b=c\)
3.
\(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\)
\(\Rightarrow\left(\dfrac{x^2-yz}{a}\right)^2=\left(\dfrac{y^2-xz}{b}\right)\left(\dfrac{z^2-xy}{c}\right)=\dfrac{\left(x^2-yz\right)^2-\left(y^2-xz\right)\left(z^2-xy\right)}{a^2-bc}\)
\(=\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}\)
Tương tự:
\(\left(\dfrac{y^2-xz}{b}\right)^2=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}\)
\(\left(\dfrac{z^2-xy}{c}\right)^2=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)
\(\Rightarrow\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)
\(\Rightarrow\dfrac{x}{a^2-bc}=\dfrac{y}{b^2-ac}=\dfrac{z}{c^2-ab}\Rightarrowđpcm\)
Cho M= \(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+x^2\)
M the a,b,c biết \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c\)
Ai giups mk vs ai nhanh mk sẽ vote nha
Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)
M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)
= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)
= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac
= 4x\(^2\)-2x(a+b+c)+ab+bc+ca
Thay 2x=a+b+c,ta được:
M= 4x\(^2\)-2x.2c+ab+bc+ca
M= 4x\(^2\)-4x\(^2\)+ab+bc+ca
M= ab+bc+ca
1. Cho biểu thức:
A = \(x-2+\dfrac{6x^2-3x}{x^3+2x^2}+\left(\dfrac{x+1}{x^2-1}+\dfrac{2}{x+1}-\dfrac{3}{x}\right):\dfrac{x+2}{x^2-1}\)
a) Rút gọn A.
b) Tìm x sao cho A nhận giá trị âm.
2. Giải phương trinh: \(\dfrac{a+b-x}{c}+\dfrac{b+c-x}{a}+\dfrac{a+c-x}{b}=1-\dfrac{4x}{a+b+c}\) với \(a,b,c\ne0\); \(a+b+c\ne0\); \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne\dfrac{4}{a+b+c}\) và x là ẩn số.
3. Giải bất phương trình: \(3x^3+4x^2+5x-6>0\).
4. Tìm x sao cho: 2 < x < 3 và \(2\left|x\right|-3\left|x-2\right|+4\left|x-3\right|=5\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)