a) 33 . 3 + (x-7)=60
b) 3 . x2 = 96
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\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
a)
\(\frac{20}{11}.\frac{33}{23}.\frac{69}{180}\)
\(=\left(\frac{20}{11}.\frac{69}{180}\right).\frac{33}{23}\)
\(=\frac{23}{33}.\frac{33}{23}=1\)
b) 0,8 x 96 + 0,8 x 2 x 2
=0,8 x ( 96+2+2)
=0,8 x 100
= 80
a) 20/11 x 33/23 x 69/180 = 60/23 x 69/180 = 1
b) 0,8 x 96 + 0,8 x 2 x 2 = 0,8 x ( 96 + 2 ) x 2 = 0,8 x 98 x 2 = 78,4 x 2 = 156,8
`A)2/3=x/60`
`=>40/60=x/60`
`=>x=40`
`B)-1/2=y/18`
`=>-9/18=y/18`
`=>y=-9`
`C)3/x=y/35=-36/84`
Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`
`=>3/x=-3/7`
`=>x=-7`
`y/35=-3/7=-15/35`
`=>y=-15`
`D)7/x=y/27=-42/54`
Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`
`=>7/x=-7/9`
`=>x=-9`
`y/27=-7/9=-21/27`
`=>y=-21`
A/
12= 22.3
60=22.3.5
ƯCLN(12,60)=22.3.5=60
B/
24=23.3
88=23.11
ƯCLN(24,88)=23.3.11=264
C/
96=25.3
224= 25.7
ƯCLN(96,224)=25.3.7=672
D/
34=2.17
96=25.3
ƯCLN(34,96)=25.3.17=1632
Chúc bạn học tốt
a) 3x2 – 7x + 2
\(=3x^2-6x-x+2\)
\(=\left(3x^2-6x\right)-\left(x-2\right)\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) a(x2 + 1) – x(a2 + 1)
\(=ax^2+a-\left(a^2x+x\right)\)
\(=a\left(x^2+1\right)-x\left(a^2+1\right)\)
.......?
a) Ta có: \(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=x^2a+a-a^2x-x\)
\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)
\(=xa\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(xa-1\right)\)
c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)
\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)
\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)
\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)
\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)
Tìm x:
a, 96 - 3 . ( x + 1 ) = 42
3 . ( x + 1 ) = 96 - 42
3 . ( x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 -1
x = 17
b, 12x - 33 = 32.33
12x - 33 =9 . 27
12x - 33 = 243
12x = 243 + 33
12x = 276
x=276:12
x=23
a, 96 - 3 . ( x + 1 ) = 42
3 . ( x + 1 ) = 96 - 42
3 . ( x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 - 1
x = 17
b, 12x - 33 = 32 . 33
12x - 33 = 9 . 27
12x - 33 = 243
12x = 243 + 33
12x = 276
x = 276 : 12
x = 23
a) \(\frac{x+3}{x+7}=\frac{96}{108}\)
\(\frac{x+3}{x+7}=\frac{8}{9}\)
\(8\left(x+7\right)=9\left(x+3\right)\)
\(8x+56=9x+27\)
\(9x-8x=56-27\)
\(x=29\)
b) \(\frac{4x+...}{-18}=-\frac{1}{6}\)(ĐỀ HÌNH NHƯ BỊ THIẾU)
\(6\left(4x+...\right)=\left(-1\right)\left(-18\right)\)
\(6\left(4x+...\right)=18\)
\(4x+...=18:6\)
\(4x+...=3\)
Bạn từ điền tiếp phần bị thiếu nhé
c) \(\frac{6}{3y-1}=\frac{33}{11}\)
\(\frac{6}{3y-1}=3\)
\(3y-1=\frac{6}{3}\)
\(3y-1=2\)
\(3y=2+1\)
\(3y=3\)
\(y=\frac{3}{3}=1\)
d) \(\frac{5z-3}{3}=\frac{28}{12}\)
\(\frac{5z-3}{3}=\frac{7}{3}\)
\(3\left(5z-3\right)=7.3\)
\(15z-9=21\)
\(15z=21+9\)
\(15z=30\)
\(z=\frac{30}{15}=2\)
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100