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bạn vào loigiaihay rồi chọn toán lớp 8 rồi chọn đẳng thức đáng nhớ
dễ mà áp dụng hết hằng đẳng thức nếu bạn thuộc hằng đẳng thức mik chỉ làm mỗi bài 1 ý nha xong dựa vô mà làm
\(1a.\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2\)
\(=4y^2+12xy+9y^2\)
\(2a.x^2-6x+9\)
\(=x^2-2.x.3+3^2\)
\(=\left(x-3\right)^2\)
a) Ta có: \(\left(x-3\right)^3\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(=x^3-9x^2+27x^2-27\)
b) Ta có: \(\left(2x-3\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)
\(=8x^3-36x^2+54x-27\)
c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)
\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
d) Ta có: \(\left(x^2-2\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)
\(=x^6-6x^4+12x^2-8\)
e) Ta có: \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-24x^2y+36xy^2-27y^3\)
f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)
\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)
ĐKXĐ \(x\ne-2,-3,-4\)
=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)
=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)
=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)
=> (3x + 7)(x + 4) - 6(x2 + 5x + 6) = 0
=> 3x2 + 19x + 28 - 6x2 - 30x - 36 = 0
=> -3x2 - 11x - 8 = 0
=> -3x2 - 3x - 8x - 8 = 0
=> -3x(x + 1) - 8(x + 1) = 0
=> (x + 1)(-3x - 8) = 0
=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)
Vậy ...
b) Thiếu dữ liệu cuả đề
c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)
ĐKXĐ \(x\ne-2;-3\)
=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)
=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)
=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)
=> 4x2 + 29x + 52 = x + 4
=> 4x2 + 29x + 52 - x - 4 = 0
=> 4x2 + 28x + 48 = 0
=> 4(x2 + 7x + 12) = 0
=> x2 + 7x +12 = 0
=> x2 + 3x + 4x + 12 = 0
=> x(x + 3) + 4(x + 3) = 0
=> (x + 3)(x + 4) = 0
=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
Mà \(x\ne-2,-3\)nên x = -3 loại
Vậy x = -4
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)