(X cộng 6 ) . (X cộng 6 )/2 - 4x/3=0
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1
TT
1
TN
19 tháng 6 2017
a)\(\left(x^2-9\right)\left(x+2\right)=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+2\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\left(x-3\right)\left(x+2\right)-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-x-6-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-x-7=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1\pm\sqrt{29}}{2}\end{cases}}\)
b)\(x^4-6x^2+4x=0\)
\(\Leftrightarrow x\left(x^3-6x+4\right)=0\)
\(\Leftrightarrow x\left[x^3+2x^2-2x-2x^2-4x+4\right]=0\)
\(\Leftrightarrow x\left[x\left(x^2+2x-2\right)-2\left(x^2+2x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+2x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0;x=2\\x=\pm\sqrt{3}-1\end{cases}}\)
c)\(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)
Đặt \(a=\sqrt{x^2-3x+3}>0\Rightarrow a^2+3=x^2-3x+6\)
\(pt\Leftrightarrow a+\sqrt{a^2+3}=3\)\(\Leftrightarrow\sqrt{a^2+3}=3-a\)
\(\Leftrightarrow a^2+3=a^2-6a+9\)
\(\Leftrightarrow6a-6=0\Leftrightarrow6\left(a-1\right)=0\Rightarrow a=1\) (thỏa)
\(\sqrt{x^2-3x+3}=1\)\(\Rightarrow x^2-3x+3=1\)
\(\Rightarrow x^2-3x+2=0\Rightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\) (thỏa)
CC
23 tháng 1 2020
\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left[\frac{x^2}{x\left(x^2-4\right)}+\frac{-6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2}{x-2}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-1}{x-2}\)
MN
11 tháng 2 2020
Ta có : \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)
\(\Leftrightarrow x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x-1=0\)
hoặc \(x^2+x+6=0\)
\(\Leftrightarrow\) \(x=-2\)(tm)
hoặc \(x=1\)(tm)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)
TP
4
HT
11 tháng 9 2017
a) \(70-5\left(2x-3\right)=45\)
\(70-10x+15=45\)
\(-10x=45-70-15\)
\(-10x=-40\)
\(x=4\)
vay \(x=4\)
b) \(156-\left(x+61\right)=82\)
\(x+61=156-82\)
\(x+61=74\)
\(x=74-61\)
\(x=13\)
vay \(x=13\)
c) \(6\left(5x+35\right)=330\)
\(30x+210=330\)
\(30x=330-210\)
\(30x=120\)
\(x=4\)
vay \(x=4\)
d) \(936-\left(4x+24\right)=72\)
\(4x+24=936-72\)
\(4x+24=864\)
\(4x=864-24\)
\(4x=840\)
\(x=210\)
vay \(x=210\)
11 tháng 9 2017
\(70-5.\left(2x-3\right)\)\(=45\)
\(5.\left(2x-3\right)\)\(=\)\(70-45\)
\(5.\left(2x-3\right)\)\(=25\)
\(2x-3=25:5\)
\(2x-3=5\)
\(2x=8\)
\(x=8:2\)
\(x=4\)
\(156-\left(x+61\right)\)\(=82\)
\(x+61=156-82\)
\(x+61=74\)
\(x=74-61\)
\(x=13\)
\(6.\left(5x+35\right)\)\(=330\)
\(5x+35=330:6\)
\(5x+35=55\)
\(5x=55-35\)
\(5x=20\)
\(x=20:5\)
\(x=4\)
\(936-\left(4x+24\right)\)\(=72\)
\(4x+24=936-72\)
\(4x+24=864\)
\(4x=864-24\)
\(4x=840\)
\(x=840:4\)
\(x=210\)
BM
27 tháng 7 2019
6 2/7 + 7 3/5 + 8 6/9 + 9 1/4 + 2/5 + 5/7 + 1/3 x 3/4 + 1967
= 44/7 + 38/5 + 78/9 + 37/4 + 2/5 + 5/7 + 1/3 + 1967
= ( 44/7 + 5/7 ) + ( 38/5 + 2/5 ) + ( 26/3 + 1/3 ) + ( 37/4 + 3/4 ) +1967
= 7 + 8 + 9 + 10 + 1967
= 15 + 9 + 10 + 1967
= 24 + 10 + 1967
= 34 + 1967
= 2001
\(\frac{\left(x+6\right)\left(x+6\right)}{2}-\frac{4x}{3}=0\)
\(\Leftrightarrow\frac{\left(x+6\right)^2}{2}=\frac{4x}{3}\)
\(\Leftrightarrow\frac{3\left(x+6\right)^2}{6}=\frac{8x}{6}\)
\(\Leftrightarrow3\left(x+6\right)^2=8x\)
\(\Leftrightarrow3\left(x^2+12x+36\right)-8x=0\)
\(\Leftrightarrow3x^2+36x+108-8x=0\)
\(\Leftrightarrow3x^2+28x+108=0\)
=> pt vô ngiệp
\(\frac{\left(x+6\right)^2}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{x^2+12x+36}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{3x^2+36x+108}{6}-\frac{8x}{6}=0\)
\(\Rightarrow\frac{3x^2+28x+108}{6}=0\)
\(\Rightarrow3x^2+28x+108=0\)
Ta có: \(\Delta=28^2-4.3.108=-512< 0\)
Vậy pt vô nghiệm