Bài 1 Bỏ ngoặc rồi tính
a, x+8-(x+22)
b, -(x+5)+ (x+10)-5
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Xét x < -5 ta có /x-5/=- ( x-5)=x+5=5-x
Khi đó /x+5 = -( x+ 5)= -x-5
B= 10 + ( 5-x) - (-x-5)
= 10+5-x+x+5=20
Xét 5 bé hoặc bằng x<5 ta có /x-5/=-(x-5)= -x+5=5-x
và /x+5/=x+5
Khi đó B= 10+(5-x) - (x+5)=10+5-x=x-5=10-2-x
Xét x lớn hoặc bằng 5 ta có /x-5/ = x-5 và /x+5/=x+5
Khi đó B=10+/x-5/-/x+5/=10+x-5-x=0
Vậy khi x < 5 thì
B= 10-2-x
Khi x>5 thì
B=0
A ) 2/17 + 30/46 + 15/17 + 4/49 + 8/23=102/49
B ) 5/8 x ( 19/31 x 8/5 )=19/31
C ) 7/9 x 8/15 x 9/7 x 15/4=2
D ) 11/12 x 24/7 x 21/22=3
nhìu quá nên mk chỉ cho kết quả thôi nha ^-^!!!
1/ (a + a).(b + b) = 2a.2b = 4ab
2/ (a - a) . (b - b) = 0.0 = 0
1) - 7264 + ( 1543 + 7264 )
= - 7264 + 1543 + 7264
= ( - 7264 + 7264 ) + 1543
= 0 + 1543
= 1543
2) ( 144 - 97 ) - 144
= 144 - 97 - 144
= ( 144 - 144 ) - 97
= 0 - 97
= - 97
3) ( - 145 ) - ( 18 - 145 )
= -145 - 18 + 145
= ( - 145 + 145 ) - 18
= 0 - 18
= - 18
4) 111 + ( - 11 + 27 )
= 111 - 11 + 27
= 100 + 27
= 127
các câu còn lại bn tự làm nha
b: \(=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)
\(a,=-3x^3+x^2+9x^2-3x-12x+4=-3x^3+10x^2-15x+4\\ b,=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3x}{x\left(x+5\right)}=\dfrac{3}{x+5}\)
a: -3x+4+5x=-10-x
=>2x+4=-x-10
=>3x=-14
hay x=-14/3
b: \(-x+1=-3x-8\)
=>-x+3x=-8-1
=>2x=-9
hay x=-9/2
c: \(8-\left(x-1\right)=10+\left(x+5\right)\)
=>x+15=8-x+1
=>x+15=9-x
=>2x=-6
hay x=-3
d: \(100+\left(x+7\right)-\left(-x+3\right)=8+\left(x+100\right)\)
=>x+7+x-3=8+x
=>2x+4-x-8=0
=>x=4
Bài 1:
a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)
b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)
Bài 2:
a)\(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\Rightarrow x=5\)
b)\(\left(x-1\right)^2=25\)
\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow x=6\) hoặc \(x=-4\)
Vậy \(x=6\) hoặc \(x=-4\)
c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)
Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)
Bài 1.
a.\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b.\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow4x-x+3x=30+5+3\)
\(\Leftrightarrow6x=38\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
Bài 1:
a. $(x-8)(x^3+8)=0$
$\Rightarrow x-8=0$ hoặc $x^3+8=0$
$\Rightarrow x=8$ hoặc $x^3=-8=(-2)^3$
$\Rightarrow x=8$ hoặc $x=-2$
b.
$(4x-3)-(x+5)=3(10-x)$
$4x-3-x-5=30-3x$
$3x-8=30-3x$
$6x=38$
$x=\frac{19}{3}$
a. (13 - 135 + 49) - (13 + 49)
= 13 - 135 + 49 - 13 - 49
= (13 - 13) + (49 - 49) - 135
= 0 + 0 - 135
= -135
a) x+8-(x+22)
= x+8-x-22
= x-x+8-22
= -14
b) -(x+5)+(x+10)-5
= -x-5+x+10-5
= -x+x-5+10-5
= 0
a)x+8-(x+22)
=x+8-x-22
=x-x+8-22
=8-22
=-14
b)-(x+5)+(x+10)-5
=-x-5+x+10-5
=-x+x-5+10-5
=-5+10-5
=0
Đúng k cho mình nha