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a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)
\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)
c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)
\(=\dfrac{-x+7x-4}{3x-2}\)
\(=\dfrac{6x-4}{3x-2}=2\)
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a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
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1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
![](https://rs.olm.vn/images/avt/0.png?1311)
mình nghĩ để sai. biểu thức ở giữa phải là (3x-1)^2 mới đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\dfrac{x-2}{2}=3.\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{x-2}{2}=\dfrac{1-3x}{2}\\ \Leftrightarrow x-2=1-3x\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)
2, mik có sửa đề vì đề của bn sai
ĐKXĐ:\(x\ne\pm\dfrac{1}{3}\)
\(\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}=\dfrac{5}{1-9x^2}\\ \Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{5}{\left(1-3x\right)\left(1+3x\right)}=0\\ \Leftrightarrow\dfrac{1-6x+9x^2-1-6x-9x^2-5}{\left(1+3x\right)\left(1-3x\right)}=0\\ \Rightarrow-12x-5=0\\ \Leftrightarrow x=-\dfrac{5}{12}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x+1\right)^2+\left(3x-1\right)^2-2\left(3x-1\right)\left(3x+1\right)\)
\(=\left(3x+1\right)^2-2.\left(3x+1\right)\left(3x1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1-3x+1\right)^2\)
\(=2^2=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow\)\(2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)
đến đây tự lm nha
b) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\) (1)
ĐKXĐ: \(x\ne\pm\frac{1}{3}\)
\(\left(1\right)\)\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
\(\Rightarrow\)\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)
\(\Leftrightarrow\)\(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)
\(\Leftrightarrow\)\(-12x=12\)
\(\Leftrightarrow\)\(x=-1\) (t/m ĐKXĐ)
Vậy....
a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)
b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
\(\Leftrightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\)
\(\Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)
\(\Leftrightarrow-12x=12\)
\(\Leftrightarrow x=-1\) (thỏa mãn)
Vậy x = -1
chiu roi
ban oi $$
sao cung duoc nhe
Vy Do
tk nhe@@@@@@@@@@__@