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1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được
\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0
<=> x + 59 = 0
<=> x=-59
a) ĐKXĐ: x≠0
Ta có: \(\frac{9}{x}+2=-6\)
⇔\(\frac{9}{x}+2+6=0\)
⇔\(\frac{9}{x}+8=0\)
⇔\(\frac{9}{x}+\frac{8x}{x}=0\)
⇔9+8x=0
⇔8x=-9
hay \(x=-\frac{9}{8}\)
Vậy: \(x=-\frac{9}{8}\)
b) ĐKXĐ: x≠0;x≠-1;x≠-3
Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)
⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)
⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)
⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)
\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)
⇔\(11x^2+7x=0\)
\(\Leftrightarrow x\left(11x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)
Vậy: \(x=\frac{-7}{11}\)
c) ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)
⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)
\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(x=\frac{-1}{3}\)
ĐKXĐ: \(x\ne2;x\ne1\)
Ta có: \(\frac{4x}{x-2}-\frac{1}{x-1}=\frac{8x^2-7}{3x-6}\)
\(\Leftrightarrow\frac{4x}{x-2}-\frac{1}{x-1}-\frac{8x^2-7}{3x-6}=0\)
\(\Leftrightarrow\frac{4x\left(x-1\right)\cdot3}{\left(x-2\right)\left(x-1\right)\cdot3}-\frac{1\left(x-2\right)\cdot3}{\left(x-1\right)\left(x-2\right)\cdot3}-\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}=0\)
\(\Leftrightarrow12x\left(x-1\right)-3\left(x-2\right)-\left(8x^2-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow12x^2-12x-\left(3x-6\right)-\left(8x^3-8x^2-7x+7\right)=0\)
\(\Leftrightarrow12x^2-12x-3x+6-8x^3+8x^2+7x-7=0\)
\(\Leftrightarrow-8x^3+20x^2-8x-1=0\)
\(\frac{12x\left(x-1\right)-3x+6}{3\left(x-2\right)\left(x-1\right)}=\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}\)
tiếp theo nhân vào và khử mẫu nha bạn!
a) \(pt\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1-\frac{3x^2+16}{x^2+10}+2=0\)
\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)
\(\Leftrightarrow4-x^2=0\)(do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0,\forall x\))
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
\(KL...\)
2x(8x - 1)2(4x - 1) = 9
<=> 512x4 - 256x3 + 40x2 - 2x = 9
<=> 512x4 - 256x3 + 40x2 - 2x - 9 = 0
<=> (2x - 1)(4x + 1)(64x4 - 16x + 9) = 0
vì 64x4 - 16x + 9 khác 0 nên:
<=> 2x - 1 = 0 hoặc 4x + 1 = 0
<=> x = 1/2 hoặc x = -1/4
x=-7206932,631
tính = máy tính đó
nhớ tink cho mình nữa nha
giải ra đi