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bài 12 :
a,\(\left(x-\frac{1}{2}\right)^2=0\)
Mà: 02=0
=> \(\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
b, \(\left(x-2\right)^2=1\)
Mà : 1=12
\(\Rightarrow\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c, \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)=-2\)
Vì -8 =-23
nên ...
=> 2x =-1
=> x=0.5
d.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
cái này cũng như mấy cái trên thôi
Bài 12:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
- Nếu \(x-2=1\)
\(x=3\)
- Nếu \(x-2=-1\)
\(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow2x-1=-2\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x+\frac{1}{12}=\pm\frac{1}{4}\)
- Nếu \(x+\frac{1}{12}=\frac{1}{4}\)
\(x=\frac{1}{6}\)
- Nếu \(x+\frac{1}{12}=-\frac{1}{4}\)
\(x=-\frac{1}{3}\)
Bài 13: có người làm rồi
Bài 14:
a) \(25^3\div5^2\)
\(=\left(5^2\right)^3\div5^2\)
\(=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{8}=2\frac{1}{8}\)
a) \(\dfrac{-1}{3}\cdot2\cdot\dfrac{-1}{3}=\left(\dfrac{-1}{3}\right)^2\cdot2=\dfrac{1}{9}\cdot2=\dfrac{2}{9}\)
c) \(\dfrac{8^4}{4^4}=\left(\dfrac{8}{4}\right)^4=2^4=16\)
d) \(\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)
a,Ta có : \(\frac{x}{x}=\frac{4y}{7}\) => \(1=\frac{4y}{7}\)=> \(2x=\frac{4y}{7}\)=> 14x = 4y => 7x = 2y => \(\frac{x}{2}=\frac{y}{7}\)=> \(\frac{2x}{4}=\frac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{4}=\frac{y}{7}=\frac{2x-y}{4-7}=\frac{3}{-3}=-1\)
=> \(\hept{\begin{cases}\frac{2x}{4}=-1\\\frac{y}{7}=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-4\\y=-7\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-7\end{cases}}\)
b, \(\frac{x}{4}=\frac{y}{3}\)=> \(\frac{x^2}{16}=\frac{y^2}{9}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{16}=\frac{y^2}{9}=\frac{x^2-y^2}{16-9}=\frac{36}{7}\)
=> Từ đó suy ra x,y không thỏa mãn điều kiện
a. \(\frac{x}{x}=\frac{4y}{7}\)=> 4y = 7 => y = \(\frac{7}{4}\)
2x - y = 3 => 2x = \(\frac{19}{4}\) => x = \(\frac{19}{8}\)
b. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4}=\frac{y}{3}=\frac{x^2-y^2}{4^2-3^2}=\frac{36}{7}\)
=> x,y \(\in\varnothing\)
Ta có : \(\frac{x-5}{5x-1}=\frac{4x-10}{20x+4}\)
=> \(\frac{x-5}{5x-1}=\frac{2x-5}{10x+2}\)
=> (x - 5)(10x + 2) = (2x - 5)(5x - 1)
=> 10x2 + 2x - 50x - 10 = 10x2 - 2x - 25x + 5
=> 10x2 - 48x - 10x2 + 27x = 5 + 10
=> -21x = 15
=> x = 15 : (-21) = -5/7
Thay x = -5/7 vào \(\frac{x-5}{5x-1}=\frac{y}{3}\)
=> \(\frac{-\frac{5}{7}-5}{5.\left(-\frac{5}{7}\right)-1}=\frac{y}{3}\)
=> \(\frac{-\frac{40}{7}}{-\frac{32}{7}}=\frac{y}{3}\)
=> \(\frac{5}{4}=\frac{y}{3}\)
=> 4y = 15
=> y = 15/4
Vậy ...
Ta có: \(\frac{5}{y}=\frac{3}{x}\) => \(\frac{x}{3}=\frac{y}{5}\) => \(\frac{x^2}{9}=\frac{y^2}{25}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{9}=\frac{y^2}{25}=\frac{y^2+x^2}{25+9}=\frac{125}{34}\)
=> \(\hept{\begin{cases}\frac{x^2}{9}=\frac{125}{34}\\\frac{y^2}{25}=\frac{125}{34}\end{cases}}\) => \(\hept{\begin{cases}x^2=\frac{125}{34}.9=\frac{1125}{34}\\y^2=\frac{125}{34}.25=\frac{3125}{34}\end{cases}}\) => \(\hept{\begin{cases}x=\pm\frac{15\sqrt{170}}{34}\\y=\pm\frac{25\sqrt{170}}{34}\end{cases}}\)
a. x^2.y^2=162
ta có \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}\)=>\(\frac{x^2}{4}=\frac{y^2}{1}=\frac{z^2}{9}\)
=>\(\frac{x^2}{4}.\frac{y^2}{1}=\frac{z^4}{81}\)còn lại do đề sai :))
a/ Ta có: P(x)=0
nên 4x2 - 3x=0
do đó: 4xx-3x=0
(4x-3)x=0
Suy ra: 4x-3 = 0 hoặc x=0
=> x=\(\dfrac{3}{4}\) hoặc x=0
Vậy x=\(\dfrac{3}{4}\) hoặc x=0 là nghiệm của P(x)
b/ P(x)=0
2x2-8x=0
Nên (2x-8)x=0
=> 2x-8=0 hoặc x=0
Do đó: x=4 hoặc x=0
Vậy x=4 hoặc x=0 là nghiệm của P(x)
c/ P(x)=0
7x-2x2=0
(7-2x)x=0
Nên 7-2x=0 hoặc x = 0
Do đó: x=\(\dfrac{7}{2}\) hoặc x = 0
Vậy x=\(\dfrac{7}{2}\) hoặc x = 0 là nghiệm của P(x)
d/ Ta có: P(x)=0
nên \(\dfrac{3}{4}x-\dfrac{1}{2}x^2=0\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)x=0\)
Do đó: \(\dfrac{3}{4}-\dfrac{1}{2}x=0\) hoặc x=0
Suy ra: x= \(\dfrac{3}{2}\) hoặc x=0
Vậy x= \(\dfrac{3}{2}\) hoặc x=0 là nghiệm của P(x)
a) \(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{2}{5}+\frac{5}{7}\)
\(x=\frac{14}{35}+\frac{25}{35}=\frac{39}{35}\)
b)
\(\frac{-2}{5}x=\frac{4}{15}\)
\(x=\frac{4}{15}:-\frac{2}{5}\)
\(x=\frac{4}{15}\cdot-\frac{5}{2}=-\frac{2}{3}\)
c) \(2x\left(x-\frac{1}{7}\right)=2x^2-\frac{2x}{7}\)
d) \(\frac{1}{2}+\frac{3}{4}x=\frac{1}{4}\)
\(\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{3}{4}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}\cdot\frac{4}{3}=-\frac{1}{3}\)
f) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{5}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{5}=\frac{31}{60}\)
\(x=\frac{31}{60}-\frac{2}{5}=\frac{7}{60}\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
sai
\(b.5-\left|2x-3\right|=\frac12\)
\(\left|2x-3\right|=5-\frac12\)
\(\left|2x-3\right|=\frac92\)
\(\Rightarrow\left[\begin{array}{l}2x-3=\frac92\\ 2x-3=-\frac92\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\left(\frac92+3\right):2\\ x=\left(-\frac92+3\right):2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{15}{4}\\ x=-\frac34\end{array}\right.\)
vậy \(x=\frac{15}{4}\) hoặc \(x=-\frac34\)
\(c.\left(\frac57-x\right)\cdot\frac{11}{15}=-\frac{22}{45}\)
\(\left(\frac57-x\right)=-\frac{22}{45}:\frac{11}{15}=-\frac{22}{45}\cdot\frac{15}{11}\)
\(\left(\frac57-x\right)=-\frac23\)
\(x=\frac57-\left(-\frac23\right)=\frac{29}{21}\)
vậy x\(=\frac{29}{21}\)
\(d.\left|x\right|-\frac23=\frac49\)
\(\left|x\right|=\frac49+\frac23\)
\(\left|x\right|=\frac{10}{9}\)
\(\Rightarrow\left[\begin{array}{l}x=\frac{10}{9}\\ x=-\frac{10}{9}\end{array}\right.\)
vậy x\(=\frac{10}{9}\) hoặc x = \(-\frac{10}{9}\)
\(e.3,2x+\left(-1,2\right)x=-4,9-2,7\)
\(2x=-7,6\)
\(x=-7,6:2=-3,8\)
vậy x=-3,8
g. \(\left|x\right|=9\cdot\left(-\frac13\right)^2+\frac13=\frac43\)
\(\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=-\frac43\end{array}\right.\)
vậy \(x=\frac43\) hoặc \(x=-\frac43\)
\(h.x^2=\sqrt{\left(-4\right)^2}\)
\(x^2=4\)
\(\Rightarrow x=\pm2\)
\(i.\frac34+\frac14:x=\frac25\)
\(\frac14:x=\frac25-\frac34\)
\(\frac14:x=-\frac{7}{20}\)
\(x=\frac14:\left(-\frac{7}{20}\right)=-\frac57\)
\(j.\left(x-3^2\right)^3=\left(3^3\right)^2\)
\(\left(x-3^2\right)^3=\left(3^2\right)^3\)
\(\Rightarrow x-3^2=3^2\)
\(x-9=9\)
⇒ x = 18
\(k.\frac{x}{7}+\left(-\frac37\right)^2=\frac27:\frac43\)
\(\frac{x}{7}+\left(-\frac37\right)^2=\frac{3}{14}\)
\(\frac{x}{7}=\frac{3}{14}-\left(-\frac37\right)^2=\frac{3}{98}\)
⇒ 21 = 98x
⇒ x = \(\frac{21}{98}=\frac{3}{14}\)