Tim x biet:
\(\frac{x-ab}{a+b}=\frac{x-bc}{b+c}=\frac{x-ca}{c+a}=a+b+c\\ \) (a khac -b ; b khac -c ; c khac -a)
Những câu hỏi liên quan
biet x=\(\frac{a}{b+c}\)=\(\frac{b}{c+a}\)=\(\frac{c}{a+b}\)
tim x trong 2 truong hop sau
a] a+b+c=0
b] a+b+c khac 0
a] x= a/b+c=b/c+a=c/a+b=a+b+c/b+c+c+a+a+b=0
=> x=0
b]
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Giai va bien luan cac phuong trinh sau:
1. \(\frac{a+b-x}{c}+\frac{a+c-x}{b}+\frac{b+c-x}{a}+\frac{4x}{a+b+c}=1\)
(an x) voi dk; a,b,b khac 0 va a+b+c khac 0
2.\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
(an x) voi dk: a,b,c khac 0
3, \(\frac{mx+3}{6}+\frac{m^2-1}{2}=\frac{x+5}{10}+\frac{2}{5}\left(x+m^2+1\right)\)
(an x)
a+b+c=\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}\) tim a,b,c khac 0
ta có:(a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)-abcgeleft(a+b+cright)left(ab+bc+caright)-frac{1}{9}left(a+b+cright)left(ab+bc+caright)frac{8}{9}left(a+b+cright)left(ab+bc+caright)frac{x}{x+yz}+frac{y}{y+zx}+frac{z}{z+xy}frac{x}{left(x+yright)left(x+zright)}+frac{y}{left(y+xright)left(y+zright)}+frac{z}{left(z+xright)left(z+yright)}frac{2left(xy+yz+zxright)}{left(x+yright)left(y+zright)left(z+xright)}lefrac{9}{4left(xy+yz+zxright)}frac{9}{4}
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ta có:
(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+x\right)\left(y+z\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\frac{9}{4\left(xy+yz+zx\right)}=\frac{9}{4}\)
Giải phương trình
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
b)\(\frac{1}{2}\left(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)
\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
b)đề bài như trên
\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
\(a,\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(a,\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}-\frac{4a+4b+4c-4x}{a+b+c}=0\)
\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\right)=0\)
\(\Leftrightarrow a+b+c-x=0\)Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\ne0\)
\(\Leftrightarrow x=a+b+c\)
Vậy phương trình có nghiệm \(x=a+b+c\)
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Giải pt ẩn x :
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
Giúp Salahhhh vớiii <3
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)
\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)
\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)
\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)
\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow x=a+b+c\)
Vậy x = a + b + c
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)
\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)
\(-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c\right)}{a+b+c}-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c-x\right)}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
\(\Rightarrow\left(a+b+c-x\right)=0\)hoặc \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
+) Nếu \(\Rightarrow\left(a+b+c-x\right)=0\)thì x = a + b + c
+) Nếu \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)thì x thỏa mãn với mọi số
iu cậu quáaa <3
Bai 1: Tim ab ; biet ab = b x 9
Bai 2 : Tim ab, biet :
a) ab x 5 = 2ab b) ab = 4ab / 9.
Bai 3 : Tim ab, biet : a 2 = b x 5
Bai 4 : Tim cac so ab biet :
a) a x3 = bx9 b) a x4 = b x6.
Bai 5 :
a) Tim abc, biet a , b , c khac nhau va : a x bc = 91.
b) Tim a , b , c khac nhau biet aa x bc = 1001
Bài 1 Rút gọn biểu thức frac{left(x+frac{1}{x^4}right)-left(x^4+frac{1}{x^4}right)-2}{left(x+frac{1}{x}right)^4+x^2+frac{1}{x^2}}.frac{x^4+1999x^2+1}{2x^2}Bài 2: Cho a,b,c thoả mãnfrac{a^3}{a^2+ab+b^2}+frac{b^3}{b^2+bc+c^2}+frac{c^2}{c^2+ca+a^2}1006tính giá trị biểu thứcMfrac{a^3+b^3}{a^2+ab+b^2}+frac{b^3+c^3}{b^2+bc+c^2}+frac{c^3+a^3}{c^2+ca+a^2}
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Bài 1 Rút gọn biểu thức
\(\frac{\left(x+\frac{1}{x^4}\right)-\left(x^4+\frac{1}{x^4}\right)-2}{\left(x+\frac{1}{x}\right)^4+x^2+\frac{1}{x^2}}.\frac{x^4+1999x^2+1}{2x^2}\)
Bài 2: Cho a,b,c thoả mãn
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^2}{c^2+ca+a^2}=1006\)
tính giá trị biểu thức
M=\(\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ca+a^2}\)
cho a,b,x dương và ab + bc + ca = abc chứng minh:
\(\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}\ge\frac{a+b+c}{4}\)
\(A=\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
\(=\frac{a^3}{a^2+ab+bc+ca}+\frac{b^3}{b^2+ab+bc+ca}+\frac{c^3}{c^2+ab+bc+ca}\)
\(=\frac{a^3}{\left(a+b\right)\left(c+a\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(c+a\right)\left(b+c\right)}\)
đến đây áp dụng cô si 3 số là đc
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