a) 1/5 - x = 40%
b) 52 - ( x - 3/4 )2 = 42
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Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
a: \(=12x^2-9x-12x^2-10x+6x+5=-13x+5\)
b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+3x\)
c: \(=x^3-3x^2+3x-1+x^3+8+3\left(x^2-16\right)\)
\(=2x^3-3x^2+3x+7+3x^2-48=2x^3+3x-41\)
d: \(=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\)
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
ĐK : 6x \(\ge0\Rightarrow x\ge0\)
Khi đó |x + 1| = x + 1
|x + 2| = x +2
|x + 3| = x +3
|x + 4| = x + 4
|x + 5| = x +5
Khi đó |x + 1| + |x + 2| + |x + 3| + |x + 4| + |x + 5| = 6x
<=> x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 6x
<=> 5x + 15 = 6x
<=> x = 15 (tm)
Vậy x = 15
b) 3x + 2 - 3x + 1 - 3x = 15.340
=> 3x(32 - 3 - 1) = 15.340
<=> 3x . 5 = 15.340
<=> 3x = 341
<=> x = 41
Vậy x = 41
a,vì /x+1/,/x+2/,/x+3/,/x+4/,/x+5/\(\ge\)0 mà /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=6x suy ra x>0
nên /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=x+1+x+2+x+3+x+4+x+5=6x ( giải thích: /x/=x khi x \(\ge0\))
suy ra 5x+21=6x suy ra x=21
b, \(3^{x+2}-3^{x+1}-3^x=15.3^{40}\)
suy ra \(3^x\left(9-3-1\right)=5.3^{41}\)
suy ra \(3^x.5=5.3^{41}\Rightarrow x=41\)
a)\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}\left(x+2\right)\)
<=>\(\frac{1}{5}x-\frac{1}{3}=\frac{2}{4}x+1\)
<=>\(-\frac{3}{10}x=\frac{4}{3}\)
<=>\(x=-\frac{40}{9}\)
b)\(\frac{5}{4}\left(x-3\right)=4+\frac{3}{2}x\)
<=>\(\frac{5}{4}x-\frac{15}{4}=4+\frac{3}{2}x\)
<=>\(-\frac{1}{4}x=\frac{31}{4}\)
<=>\(x=-31\)
c)\(\frac{5}{4}\left(x-3\right)=\frac{3}{2}\left(x+4\right)\)
<=>\(\frac{5}{4}x-\frac{15}{4}=\frac{3}{2}x+6\)
<=>\(-\frac{1}{4}x=\frac{9}{4}\)
<=>x=-9
Ko bít
Vậy trả lời lm j