a, \((x^2+x)+4(x^2+x)=12\)
b, x(x-1)(x+1)(x+2)=24.
c, (x-7)(x-5)(x-4)(x-2)=72
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AH
Akai Haruma
Giáo viên
24 tháng 8 2023
1. Đặt $x^2+x=a$ thì pt trở thành:
$a^2+4a=12$
$\Leftrightarrow a^2+4a-12=0$
$\Leftrightarrow (a-2)(a+6)=0$
$\Leftrightarrow a-2=0$ hoặc $x+6=0$
$\Leftrightarrow x^2+x-2=0$ hoặc $x^2+x+6=0$
Dễ thấy $x^2+x+6=0$ vô nghiệm.
$\Rightarrow x^2+x-2=0$
$\Leftrightarrow (x-1)(x+2)=0$
$\Leftrightarrow x=1$ hoặc $x=-2$
AH
Akai Haruma
Giáo viên
24 tháng 8 2023
2.
$x(x-1)(x+1)(x+2)=24$
$\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24$
$\Leftrightarrow (x^2+x)(x^2+x-2)=24$
$\Leftrightarrow a(a-2)=24$ (đặt $x^2+x=a$)
$\Leftrightarrow a^2-2a-24=0$
$\Leftrightarrow (a+4)(a-6)=0$
$\Leftrightarrow a+4=0$ hoặc $a-6=0$
$\Leftrightarrow x^2+x+4=0$ hoặc $x^2+x-6=0$
Nếu $x^2+x+4=0$
$\Leftrightarrow (x+\frac{1}{2})^2=\frac{1}{4}-4<0$ (vô lý - loại)
Nếu $x^2+x-6=0$
$\Leftrightarrow (x-2)(x+3)=0$
$\Leftrightarrow x-2=0$ hoặc $x+3=0$
$\Leftrightarrow x=2$ hoặc $x=-3$
NT
3
2 tháng 3 2019
\(a)\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
Đặt \(t=x^2+x\left(t\ge0\right)\)
\(\Leftrightarrow t^2+4t-12=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
Với \(t=2\Rightarrow x^2+x=2\Rightarrow x^2-x-2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Với \(t=-6\Rightarrow x^2+x=-6\Rightarrow x^2+x+6=0\Rightarrow x\notin\)
Vậy...
KD
24 tháng 2 2019
Nguyễn TrươngNguyễn Việt LâmNguyenTruong Viet TruongKhôi BùiAkai HarumaÁnh LêDƯƠNG PHAN KHÁNH DƯƠNGPhùng Tuệ Minhsaint suppapong udomkaewkanjana
7 tháng 7 2023
Bài 1
a) \(5\times72\times10\times2=\left(5\times2\times10\right)\times72=100\times72=7200\)
b) \(40\times125=5\times\left(8\times125\right)=5\times1000=5000\)
c) \(16\times6\times25=4\times4\times6\times25=\left(4\times6\right)\times\left(4\times25\right)=24\times100=2400\) Bài 2:
a) \(24\times57+43\times24=24\times\left(57+43\right)=24\times100=2400\)
b) \(12\times19+80\times12+12=12\times\left(19+80+1\right)=12\times100=1200\)
c) \(\left(36\times15\times169\right)\div\left(5\times18\times13\right)\)
\(=\left(18\times2\times3\times5\times13\times13\right)\div\left(5\times18\times13\right)\)
\(=\left(2\times3\times13\right)\times\left(18\times5\times13\right)\div\left(5\times18\times13\right)\)
\(=2\times3\times13\)
\(=78\)
d) \(\left(44\times52\times60\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times11\times4\times13\times4\times15\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times4\times4\right)\times\left(11\times13\times15\right)\div\left(11\times13\times15\right)\)
\(=4\times4\times4\)
\(=64\)
Bài 3:
a) \(x-280\div35=5\times54\)
\(x-8=270\)
\(x=270+8\)
\(x=278\)
b) \(\left(x-280\right)\div35=54\div4\)
\(\left(x-280\right)\div35=\dfrac{27}{2}\)
\(x-280=\dfrac{27}{2}\times35\)
\(x-280=\dfrac{945}{2}\)
\(x=\dfrac{945}{2}+280\)
\(x=\dfrac{1505}{2}\)
c) \(\left(x-128+20\right)\div192=0\)
\(x-128+20=0\times192\)
\(x-128+20=0\)
\(x-128=0-20\)
\(x-128=-20\)
\(x=-20+128\)
\(x=108\)
d) \(4\times\left(x+200\right)=460+85\times4\)
\(4\times\left(x+200\right)=460+340\)
\(4\times\left(x+200\right)=800\)
\(x+200=800\div4\)
\(x+200=200\)
\(x=200-200\)
\(x=0\)
Bài 4:
a) \(\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{8}{11}+\dfrac{19}{11}=\dfrac{27}{11}\)
c) \(\dfrac{3}{8}+\dfrac{5}{12}=\dfrac{9}{24}+\dfrac{10}{24}=\dfrac{19}{24}\)
d) \(\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)
Bài 5:
a) \(x-\dfrac{6}{7}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{6}{7}\)
\(x=\dfrac{47}{14}\)
b) \(\dfrac{12}{7}\div x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\dfrac{12}{7}\div x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\dfrac{12}{7}\div x=\dfrac{11}{15}\)
\(x=\dfrac{12}{7}\div\dfrac{11}{15}\)
\(x=\dfrac{180}{77}\)
7 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`5 \times 72 \times 10 \times 2`
`= 5 \times 2 \times 10 \times 72`
`= 10 \times 10 \times 72`
`= 100 \times 72`
`= 7200`
`b)`
`40 \times 125`
`= 4 \times 10 \times 25 \times 5`
`= (5 \times 10) \times (4 \times 25)`
`= 50 \times 100`
`= 5000`
`c)`
`4 \times 2021 \times 25`
`= (4 \times 25) \times 2021`
`= 100 \times 2021`
`= 202100`
`d)`
`16 \times 6 \times 25`
`= 4 \times 4 \times 6 \times 25`
`= (4 \times 25) \times 4 \times 6`
`= 100 \times 24`
`= 2400`
`2,`
`a)`
`24 \times 57 + 43 \times 24`
`= 24 \times (57+43)`
`= 24 \times 100`
`= 2400`
`b)`
`12 \times 19 + 80 \times 12 +12`
`= 12 \times (19 + 80 + 1)`
`= 12 \times 100`
`= 1200`
`c)`
`(36 \times 15 \times 169) \div (5 \times 18 \times 13)`
`= 36 \times 15 \times 169 \div 5 \div 18 \div 13`
`= 6 \times 6 \times 3 \times 5 \times 13 \times 13 \div 5 \div 3 \times 6 \div 13`
`= (6 \div 6) \times (3 \div 3) \times (5 \div 5) \times (13 \div 13) \times 6 \times 13`
`= 6 \times 13`
`= 78`
`d)`
`(44 \times 52 \times 60) \div ( 11 \times 13 \times 15)`
`= 44 \times 52 \times 60 \div 11 \div 13 \div 15`
`= 4 \times 11 \times 13 \times 4 \times 15 \times 4 \div 11 \div 13 \div 15`
`= (11 \div 11) \times (13 \div 13) \times (15 \div 15) \times 4 \times 4 \times`
`= 4 \times 4 \times 4`
`= 64`
`3,`
`a)`
`x - 280 \div 35 = 5 \times 54`
`x - 8 = 270`
`x = 270 + 8`
`x = 278`
`b)`
`(x - 280) \div 35 = 54 \div 4`
`(x - 280) \div 35 = 13,5`
`x - 280 = 13,5 \times 35`
`x - 280 = 472,5`
`x = 472,5 + 280`
`x = 752,5`
`c)`
`(x - 128 + 20) \div 192 = 0`
`x - 128 + 20 = 0 \times 192`
`x - 128 + 20 = 0`
`x - 108 = 0`
`x = 0 + 108`
`x = 108`
`d)`
`4 \times (x + 200) = 460 + 85 \times 4`
`4 \times (x+200) = 460 + 340`
`4 \times (x+200) = 800`
`x + 200 = 800 \div 4`
`x + 200 = 200`
`x = 200 - 200`
`x = 0`
`4,`
`a)`
`7/12 - 5/12`
`= (7 - 5)/12`
`= 2/12`
`= 1/6`
`b)`
`8/11 + 19/11`
`= (8+19)/11`
`= 27/11`
`c)`
`3/8 + 5/12`
`= 9/24 + 10/24`
`= 19/24`
`d)`
`3/4 + 7/12`
`= 9/12 + 7/12`
`= 16/12`
`= 4/3`
`5,`
`a)`
`x - 6/7 = 5/2`
`x = 5/2 + 6/7`
`x = 47/14`
`b)`
`12/7 \div x + 2/3 = 7/5`
`12/7 \div x = 7/5 - 2/3`
`12/7 \div x = 11/15`
`x = 12/7 \div 11/15`
`x = 180/77`
`@` `\text {Kaizuu lv uuu}`
TV
12 tháng 11 2018
b; x(x-1)(x+1)(x+2)-24
=(x2+x)(x2+x-2)-24
Đặt x2+x=k khi đó k(k-2)-24=k2-2k-24
=(k2-2k+1)-25=(k-1)2-52
=(k-1-5)(k-1+5)=(k-6)(k+4)
c; (x+2)(x-2)(x2-10)-72
=(x2-4)(x2-10)-72
Đặt x2-7=k khi đó (k-3)(k+3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-7-9)(x2-7+9)
=(x2-16)(x2+2)=(x-4)(x+4)(x2+2)
d; (x-7)(x-5)(x-4)(x-2)-72
=(x2-9x+14)(x2-9x+8)-72
Đặt x2-9x+11=k khi đó (k+3)(k-3)-72=k2-9-72
=k2-81=(k-9)(k+9)=(x2-9x+11-9)(x2-9x+11+9)
=(x2-9x+2)(x2-9x+20)
=(x2-9x+2)(x2-4x-5x+20)
=(x2-9x+2)(x-4)(x-5)
Câu a:
\((x^2+x)^2+4(x^2+x)=12\)
\(\Leftrightarrow (x^2+x)^2+4(x^2+x)+4=16\)
\(\Leftrightarrow (x^2+x+2)^2=16\)
\(\Rightarrow \left[\begin{matrix} x^2+x+2=4\\ x^2+x+2=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2+x-2=0\\ x^2+x+6=0\end{matrix}\right.\)
Với \(x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x^2+x+6=0\Leftrightarrow (x^2+x+\frac{1}{4})+\frac{23}{4}=0\)
\(\Leftrightarrow (x+\frac{1}{2})^2=\frac{-23}{4}<0\) (vô lý- loại)
Vậy \(x\in \left\{-2;1\right\}\)
Câu b:
\(x(x-1)(x+1)(x+2)=24\)
\(\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24\)
\(\Leftrightarrow (x^2+x)(x^2+x-2)=24\)
\(\Leftrightarrow a(a-2)=24\) (đặt \(x^2+x=a\) )
\(\Leftrightarrow a^2-2a-24=0\)
\(\Leftrightarrow (a-6)(a+4)=0\Rightarrow \left[\begin{matrix} a-6=0\\ a+4=0\end{matrix}\right.\)
Nếu \(a-6=0\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)
Nếu \(a+4=0\Leftrightarrow x^2+x+4=0\Leftrightarrow (x+\frac{1}{2})^2=\frac{-15}{4}<0\) (vô lý)
Vậy............