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a)
(2x+1)2=25
=> \(\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}2x=4\\2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
d)
(x-1)3=-125
=> x-1=-5
=> x=-4
còn câu b và c bạn viết đề rõ hơn nha
a) (2x+1)2=25
(2x+1)2= (+-5)2
=> 2x+1 = 5 hoặc 2x + 1 = -5
2x = 4 hoặc 2x = -6
x= 2 hoặc x=-3
b) (x-1)3=-125
(x-1)3= (-5)3
=> x-1 = -5
x= -4
c) 2x+2-2x=96
2x.22 - 2x = 96
2x( 4-1) = 96
2x = 96 : 3
2x= 32
2x = 25
=> x= 5
d) 7x+2+2.7x-1=345
7x-1 . 73 + 2.7x-1=345
7x-1( 73 +2) = 345
7x-1 . 345 = 345
7x-1 =1
=> x-1 = 0
=> x= 1
a.\(3^{-2}.3^2.27^x=\frac{1}{3}\)
\(\Rightarrow3^{-2+2}.\left(3^3\right)^x=\frac{1}{3}\)
\(\Rightarrow3^0.3^{3x}=3^{-1}\)
\(\Rightarrow3^{3x}=3^{-1}\)
=> 3x=-1
=> x=\(-\frac{1}{3}\)
b.\(7^{x+2}+2.7^{x-1}=345\)
\(\Rightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Rightarrow7^{x-1}.345=345\)
=> 7x-1=345 : 345
=> 7x-1=1
=> 7x-1=70
=> x-1=0
Vậy x=1.
c.\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)\in\left\{-1;0;1\right\}\)
=> 2x-1=-1 hoặc 2x-1=0 hoặc 2x-1=1
=> 2x=0 hoặc 2x=1 hoặc 2x=2
=> x=0 hoặc x=\(\frac{1}{2}\) hoặc x=1
Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)
a) \(\left|x\left(x-7\right)\right|=x\)
\(\Rightarrow\orbr{\begin{cases}x\left(x-7\right)=x\\x\left(x-7\right)=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)
b) \(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow x=2,8\)
\(a.\)\(\left|x.\left(x-7\right)\right|=x\)( Đk: \(x\ge0\))
\(\Leftrightarrow\orbr{\begin{cases}x.\left(x-7\right)=x\\x.\left(x-7\right)=-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=x:x\\x-7=-x:x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1+7\\x=-1+7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
\(b.\)\(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)( Đk: \(5x\ge0\Leftrightarrow x\ge0\))
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(-1,1+1,2+1,3+1,4\right)=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow2,8=5x-4x\)
\(\Leftrightarrow x=2,8\)
\(c.\)\(7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.7^{x+3}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^{x+3}+2\right)=345\)
\(......................\)
Đến đây mk ko bt làm nữa, tự lm nhé !
a: =>|1/3x|=3:2,7=10/9
=>1/3=10/9 hoặc 1/3x=-10/9
=>x=10/3 hoặc x=-10/3
b: =>2|2x-1|=19-7=12
=>|2x-1|=6
=>2x-1=6 hoặc 2x-1=-6
=>2x=7 hoặc 2x=-5
=>x=7/2 hoặc x=-5/2
c: |x|>2
=>x>2 hoặc x<-2
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
\(a,A\left(x\right)=2x+3\)
Có \(2x+3=0\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy \(-\frac{3}{2}\)là 1 nghiệm của đa thức A(x)
\(b,B\left(x\right)=4x^2-25\)
\(\Rightarrow B\left(x\right)=\left(2x\right)^2-25\)
Có \(B\left(x\right)=0\Rightarrow\left(2x\right)^2-25=0\)
\(\Rightarrow\left(2x\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x=5\\2x=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)
Vậy -5/2 là 1 nghiệm của B(x)
\(c,C\left(x\right)=x^2-7\)
Có \(C\left(x\right)=0\Leftrightarrow x^2-7=0\)
\(\Rightarrow x^2=7\)
\(\Rightarrow x=\orbr{\begin{cases}\sqrt{7}\\-\sqrt{7}\end{cases}}\)
Vậy \(\sqrt{7};-\sqrt{7}\)là 2 nghiệm của C(x)
\(d,D\left(x\right)=x\left(1-2x\right)+\left(2x^2-x+4\right)\)
\(D\left(x\right)=x-2x^2+2x^2-x+4\)
\(D\left(x\right)=4\)
Vậy D(x) vô nghiệm
+) Ta có: A(x) = 2x + 3 = 0
(=) 2x = -3
(=) x = \(\frac{-3}{2}\).
+) Ta có: B(x) = 4x2 -25 = 0
(=) 4x2 = 25
(=) (2x)2 = 52
=> 2x = 5
(=) x = \(\frac{5}{2}\).
a) \(\left(2x+1\right)^2=25\)
=> \(2x+1=5\) và \(2x+1=-5\)
=> \(2x=5-1=4\) và \(2x=-5-1=-6\)
=> \(x=4:2=2\) và \(x=-6:2=-3\)
b) \(\left(x-1\right)^3=-125\)
=> \(x-1=-5\Rightarrow x=-5+1=-4\)
c) \(2^{x+2}-2^x=96\)
=> \(2^x\cdot2^2-2^x\cdot1=96\)
=> \(2^x\left(2^2-1\right)=96\)
=> \(2^x\cdot3=96\Rightarrow2^x=96:2=32\)
=> \(x=5\)
d) \(7^{x+2}+2\cdot7^{x-1}=345\)
=> \(7^x\cdot7^2+2\cdot7^x:7=345\)
=> \(7^x\cdot7^2+2\cdot7^x\cdot\frac{1}{7}=345\)
=> \(7^x\cdot\left(7^2+2\cdot\frac{1}{7}\right)=345\)
=> \(7^x\cdot\frac{345}{7}=345\)
=> \(7^x=345:\frac{345}{7}=7\)
=> \(x=1\)
\(\left(2x+1\right)^2=25\)
\(\left(2x+1\right)^2=5^2=\left(-5\right)^2\)
\(TH1:\left(2x+1\right)^2=5^2\)
\(2x+1=5\)
\(x=\left(5-1\right):2\)
\(x=4\)
\(TH2:\left(2x+1\right)^2=\left(-5\right)^2\)
\(2x+1=-5\)
\(x=\left[\left(-5\right)-1\right]:2\)
\(x=-3\)
Vậy x=2 hoặc x= -3