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a, Ta có : \(f\left(x\right)-g\left(x\right)=h\left(x\right)\)hay
\(4x^2+3x+1-3x^2+2x-1=h\left(x\right)\)
\(\Rightarrow h\left(x\right)=x^2+5x\)
b, Đặt \(h\left(x\right)=x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy nghiệm của đa thức h(x) là x = -5 ; x = 0
Đặt \(k\left(x\right)=7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow7\left(x^2+2x+3x+6\right)=0\Leftrightarrow7\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)
Vậy nghiệm của đa thức k(x) là x = -3 ; x = -2
xin lỗi mọi người 1 tý nha cái phần c) ý ạ đề thì vậy như thế nhưng có cái ở phần biểu thức ở dưới ý là
\(\left(\frac{3^2}{6}-81\right)^3\) chuyển thành \(\left(\frac{3^3}{6}81\right)^3\)
bị sai mỗi thế thôi ạ mọi người giúp em với ạ
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
a/ \(\left|3x-1\right|=\left|5-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)
Vậy ......
b/ \(\left|x+2\right|-\left|x+7\right|=0\)
\(\Leftrightarrow\left|x+2\right|=\left|x+7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x+7\\x+2=-x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x=7-2\\x+x=-7-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\2x=-9\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{9}{2}\)
Vậy ...............
c/ \(\left|2x-1\right|+x=2\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2-x\\2x-1=-2+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+x=2+1\\2x-x=-2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy ..
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
\(\left(3-\frac{1}{2}x\right)\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\left|x+\frac{3}{4}\right|=\frac{5}{6}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=6\\x=\frac{1}{12}\\x=\frac{-19}{12}\end{cases}}\)
\(\left(3-\frac{1}{2}x\right)\cdot\left(\left|x+\frac{3}{4}\right|-\frac{5}{6}\right)=0\)
\(\Rightarrow\hept{\begin{cases}3-\frac{1}{2}x=0\\\left|x+\frac{3}{4}\right|-\frac{5}{6}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=6\\x+\frac{3}{4}=\pm\frac{5}{6}\end{cases}}\)
Ta có
\(x+\frac{3}{4}=\pm\frac{5}{6}\)
\(\hept{\begin{cases}x+\frac{3}{4}=\frac{5}{6}\\x+\frac{3}{4}=-\frac{5}{6}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{12}\\x=-\frac{19}{12}\end{cases}}}\)
Vậy \(x\in\left\{3;\frac{1}{2};-\frac{19}{12}\right\}\)
a) 3.x-14 =81
3.x = 81+1
3.x = 82
x = 82:3
x=82/3
Vậy x = 82/3
b) (x+1)5= -32
(x+1)5= (-2)5
x+1 = -2
x = -2-1
x = -3
Vậy x = -3