12x+4/21-x-3/3=3(x-2)/7
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\(a,\dfrac{1}{12}\times\dfrac{4}{5}=\dfrac{1\times4}{12\times5}=\dfrac{1}{15}\)
\(b,\dfrac{40}{7}\times\dfrac{21}{5}=\dfrac{40\times21}{7\times5}=\dfrac{24}{1}=24\)
\(c,\dfrac{9}{5}\div\dfrac{4}{7}=\dfrac{9}{5}\times\dfrac{7}{4}=\dfrac{9\times7}{5\times4}=\dfrac{63}{20}\)
\(d,\dfrac{11}{24}\div\dfrac{44}{3}=\dfrac{11}{24}\times\dfrac{3}{44}=\dfrac{11\times3}{24\times44}=\dfrac{1}{32}\)
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
\(d,x\left(x-3\right)-7x+21=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
\(a,2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
\(c,4x^2+12x+9=0\)
\(\Leftrightarrow4x^2+6x+6x+9=0\)
\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ
7 x 2 = 14 7 x 4 = 28 7 x 6 = 42 7 x 3 = 21
2 x 7 = 14 4 x 7 = 28 6 x 7 = 42 3 x 7 = 21
14 : 7 = 2 28 : 7 = 4 42 : 7 = 6 21 : 7 = 3
14 : 2 = 7 28 : 4 = 7 42 : 6 = 7 21 : 3 = 7
a/ 128 . 3 . (x + 4) = 23
=> 384. (x + 4) = 23
=> x + 4 = 23/384
=> x = -1513/384
b/ 100 - 7 .(x + 5) = 58
=> 7. (x + 5) = 42
=> x + 5 = 6
=> x = 1
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
(12x + 4)/21 - (x - 3)/3 = (3(x - 2))/7
<=> (4(3x + 1))/21 - (x - 3)/3 = (3(x - 2))/7
<=> 4(3x + 1) - 7(x - 3) = 9(x - 2)
<=> 12x + 4 - 7x + 21 = 9x - 18
<=> 5x + 25 = 9x - 18
<=> 5x + 25 - 9x = -18
<=> -4x + 25 = -18
<=> -4x = -18 - 25
<=> -4x = -43
<=> x = 43/4