
a)
20
9...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(x^8+x^4+1\) \(=\left(x^8+2x^4+1\right)-x^4\) \(=\left(x^4+1\right)^2-x^4\) \(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\) \(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\) \(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\) \(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\) \(\frac{-5}{9}x+1=\frac{2}{3}x-10\) \(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\) \(-5x+9=6x-90\) \(-5x-6x=-90-9\) \(-11x=-99\) \(x=\frac{-99}{-11}=9\) b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\) \(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\) \(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\) \(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\) x=30 Chúc bạn học tốt!! \(a.\left(-15\right)^5:\left(-15\right)^3\) \(=\left(-15\right)^{5-3}\) \(=\left(-15\right)^2\) \(b.\left(\frac{5}{8}\right)^{10}:\left(\frac{5}{8}\right)^3\) \(=\left(\frac{5}{8}\right)^{10-3}\) \(=\left(\frac{5}{8}\right)^7\) \(c.x^{10}:x^7\) \(=x^{10-7}\) \(=x^3\) \(d.y^5:y^4\) \(=y^{5-4}\) \(=y^1=y\) \(e.\left(-x.y^2.z\right)^3:\left(-x.y^2.z\right)^3\) \(=\left(-x.y^2.z\right)^{3-3}\) \(=\left(-x.y^2.z\right)^0=1\) câu a nè = (4x-1)(2x-3) câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx) Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\) \(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\) Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\) Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai) b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\) => k2 \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\) => k = -2;2 + k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\) \(\frac{z}{7}=2\Rightarrow z=14\) + k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\) \(\frac{z}{7}=2\Rightarrow z=-14\) Vậy................................ 1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\) \(MC:12\) Quy đồng : \(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\) \(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\) \(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\) \(\Leftrightarrow6x+9-3x=-4-9+16\) \(\Leftrightarrow-7x=3\) \(\Leftrightarrow x=\frac{-3}{7}\) 2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\) \(MC:20\) Quy đồng : \(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\) \(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\) \(\Leftrightarrow30x+15-20=15x-2\) \(\Leftrightarrow15x=3\) \(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
