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a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)
\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)
\(\Leftrightarrow6x+6-2x-2=5x+5-1\)
\(\Leftrightarrow6x-2x-5x=5-1-6+2\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)
\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)
\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)
\(\Leftrightarrow2x+6x=-7-5\)
\(\Leftrightarrow8x=-12\)
\(\Leftrightarrow x=-\frac{3}{2}\)
c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)
\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)
\(\Leftrightarrow\frac{x}{2}-\frac{2}{3}x+\frac{5}{6}x=-2+1-\frac{4}{3}+\frac{5}{3}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)
\(\Leftrightarrow x=-1\)
câu a mình ko biết làm
b.A=\(\frac{a}{b+c}\)=\(\frac{c}{a+b}=\frac{b}{c+a}=\frac{a+c+b}{b+c+a+b+c+a}=\frac{a+b+c}{\left(a+b+c\right)^2}=\frac{1}{2}\)
chúc bạn học tốt
Bài 5 :
a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)
=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
=> \(36x+3=0\)
=> \(x=-\frac{1}{12}\)
Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)
b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)
=> \(35x-5+60x-96+6x=0\)
=> \(101x-101=0\)
=> \(x=1\)
Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)
c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
=> \(\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(2x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)
=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)
=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)
=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)
=> \(-64x+123=0\)
=> \(x=\frac{123}{64}\)
Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
1. Tìm x, biết :
a. ( x - \(\frac{3}{4}\)) \(^2\)= 0
=> x - \(\frac{3}{4}\)= 0
=> x = 0 + \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)
=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)
=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)
=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)
=> x = \(\frac{-1}{8}\)
c. \(\frac{\left(-2\right)^x}{16}=-8\)
=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)
=> ( -2)\(^x\)= -128
=> ( -2 ) \(^x\)= ( -2) \(^7\)
=> x = 7