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\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
1) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x.1+5^x.5^2=650\)
\(\Rightarrow5^x.\left(1+5^2\right)=650\)
\(\Rightarrow5^x.26=650\)
\(\Rightarrow5^x=650:26\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Mình chỉ làm câu 1) thôi nhé.
Chúc bạn học tốt!
Bài 1 :
a. \(\left|x-\frac{1}{3}\right|< \frac{5}{2}\)
TH1 : nếu \(\left|x-\frac{1}{3}\right|>0\)
\(x-\frac{1}{3}< \frac{5}{3}\)
\(x< 2\)
TH2 : nếu \(\left|x-\frac{1}{3}\right|< 0\)
\(\frac{1}{3}-x< \frac{5}{3}\)
\(x>-\frac{4}{3}\)
Bài 2 :
a. \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2-1=0\)
\(\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
1,+) Thay x = 5 vào biểu thức A, ta có:
A = 4.52 - 5.|5| + 2.|3 - 5|
A = 4.25 - 5.5 + 2.2
A = 100 - 25 + 4
A = 75 + 4 = 79
Thay x = 3 vào biểu thức A, ta có:
A = 4.32 - 5.|3| + 2.|3 - 3|
A = 4.9 - 5.3 + 2.0
A = 36 - 15 = 21
+) Ta có: B = xy + x2y2 + x3y3 + ... + x100y100
B = xy + (xy)2 + (xy)3 + ... + (xy)100
Thay x = 1; y= -1 vào biểu thức B, ta có:
B = 1.(-1) + [1.(-1)]2 + [1.(-1)]3 + ... + [1.(-1)]100
B = -1 + 1 - 1 + ... + 1
B = 0
+) Thay x = 1 vào C, ta có:
C = 100.1100 + 99.199 + 98.198 + ... + 2.12 + 1
C = 100 + 99 + 98 + ... + 2 + 1
C = (100 + 1).[(100 - 1) : 1 + 1] : 2
C = 101.100 : 2
C = 5050
+) Thay x = 99 vào biểu thức D, ta có:
D = 9999 - 100.9998 + 100.9997 - 100.9996 + ... + 100.99 - 1
D = 9999 - (99 + 1).9998 + (99 + 1).9997 - (99 + 1).9996 + ... + (99 + 1).99 - 1
D = 9999 - 9999 - 9998 + 9998 + 9997 - 9997 - 9996 + ... + 992 + 99 - 1
D = 99 - 1 = 98
a) \(x\left(xy+1\right)+y\left(xy-1\right)-xy\left(x+y\right)\)
\(=X^2y+x+xy^2-y-x^2y-xy^2\)
\(=x-y\)
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
a)4x-1+5.4x-2=576
=> 4x-1(1+5.\(4^{-1}\))=576
=> 4x-1.\(\dfrac{9}{4}\)=576
=> 4x-1=256=44
=> x-1=4
=> x=5
b) (2x-1)6=(2x-1)8
=> (2x-1)6 - (2x-1)8=0
=> (2x-1)6(1- (2x-1)2)=0
=>\(\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.=>\left[{}\begin{matrix}2x=1\\\left(2x-1\right)^2=1hoặc\left(2x-1\right)^2=-1\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=1hoặc2x-1=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=2hoặc2x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1hoặcx=0\end{matrix}\right.\)
Vậy x\(\in\)\(\left\{\dfrac{1}{2},1,0\right\}\)
c) (2x-5)2000+(3y+4)2002 \(\le0\)
Có (2x-5)2000\(\ge\)0 với mọi x
(3y+4)2002\(\ge\)0 với mọi y
=> (2x-5)2000+(3y+4)2002 \(\ge\) 0
=> Để (2x-5)2000+(3y+4)2002 \(\le0\) thì (2x-5)2000+(3y+4)2002 =0
=> \(\left\{{}\begin{matrix}\left(2x-5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{matrix}\right.=>\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy x=\(\dfrac{5}{2}\);y=\(\dfrac{-4}{3}\)
Bài 2:
Có A=2100-299+298-...+22-2
=> 2A=2(2100-299+298-...+22-2)
=> 2A= 2101-2100+299-...+23-22
=> 2A= 2101-2100+299-...+23-22
+A= 2100-299+298-...+22-2
=> 3A= 2101-2
=> A=\(\dfrac{2^{101}-2}{3}\)