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Answer:
Câu 1:
\(\left(5x-x-\frac{1}{2}\right)2x\)
\(=\left(4x-\frac{1}{2}\right)2x\)
\(=4x.2x-\frac{1}{2}.2x\)
\(=8x^2-x\)
\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)
\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)
\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)
\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)
\(=x^4+8x^3+19x^2+24x+48\)
Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)
Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(= (x²+2xy+y²)-(x²-2xy+y²)\)
\(= x²+2xy+y²-x²+2xy-y²\)
\(= 4xy\)
\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)
Câu 2:
\(x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(x^2.\left(x-1\right)+4-4x=0\)
\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x-2=0\Rightarrow x=2\)
Trường hợp 3: \(x+2=0\Rightarrow x=-2\)
Câu 3: Bạn xem lại đề bài nhé.
a) B xác định\(\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\end{cases}}\Rightarrow x\ne\pm1\)
b) \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Mà x khác 1 nên x = 0
\(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)
\(=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}+\frac{4}{x^2-1}\)
\(=\frac{x^2-2x+1-x^2-2x-1}{\left(x+1\right)\left(x-1\right)}+\frac{4}{x^2-1}\)
\(=\frac{-4x}{\left(x+1\right)\left(x-1\right)}+\frac{4}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{-4x+4}{\left(x+1\right)\left(x-1\right)}=\frac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{-4}{x+1}\)
Thay x = 0 vào B, ta được \(P=\frac{-4}{0+1}=-4\)
Vậy P = -4 khi \(x^2-x=0\)
c) \(B=-3\Leftrightarrow\frac{-4}{x+1}=-3\Leftrightarrow x+1=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy B = -3 khi \(x=\frac{1}{3}\)
d) \(B< 0\Leftrightarrow\frac{-4}{x+1}< 0\Leftrightarrow x+1>0\Leftrightarrow x>-1\)
Vậy x > - 1 thì B < 0
tích cho cậu là ấn vào link hay là thích
Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
a) ĐKXĐ : \(x\ne\pm2\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{x^2-4}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x-23\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x^2-17x-46\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-\left(3x^2-17x-46\right)-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-3x^2+17x+46-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{2x^2+7x+6}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(2x+3\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+3}{x-2}\)
b) x2 - 1 = 0 <=> x2 = 1 <=> x = ±1
Với x = 1
\(B=\frac{2\cdot1+3}{1-2}=-5\)
Với x = -1
\(B=\frac{2\cdot\left(-1\right)+3}{\left(-1\right)-2}=-\frac{1}{3}\)
yx=10⇒x=10y
M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}M=8x2−24xy16x2−40xy=8x(x−3y)8x(2x−5y)=x−3y2x−5y
=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}=10y−3y2.10y−5y=715
Câu 2
\(\frac{x}{y}\)nha bạn