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Câu 1 :
\(a,\left(3x+2\right)^2=9x^2+12x+4.\)
\(b,\left(6a^2-b\right)^2=36a^4-12a^2b-b^2\)
\(c,\left(4x-1\right)\left(4x+1\right)=16x^2-1\)
\(d,\left(1-x\right)\left(1+x\right)\left(1+x^2\right)=\left(1-x^2\right)\left(1+x^2\right)=1-x^4\)
\(e,\left(a^2+b^2\right)\left(a^2-b^2\right)=a^4-b^4\)
\(f,\left(x^3+y^2\right)\left(x^3-y^2\right)=x^6-y^4\)
Bài 2 :
\(a,A=9x^2+42x+49=9+42+49=100.\)
\(b,B=25x^2-2xy+\frac{1}{25}y^2=\left(5x^2\right)-2.5x.\frac{1}{5}y+\left(\frac{1}{5}y\right)^2\)
\(=\left(5x-\frac{1}{5}y\right)^2=\left(-1+1\right)^2=0\)
\(c,C=4x^2-28x+49=4x^2-14x-14x+49\)
\(=2x\left(x-7\right)-7\left(x-7\right)=\left(2x-7\right)\left(x-7\right)\)
\(=\left(8-7\right)\left(4-7\right)=-3\)
Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn
Bài làm:
Bài 1
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\rightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Bài 2
a) \(25^3\div5^2=\left(5^2\right)^3\div5^2=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}\div\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}\div\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}\div\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(\frac{-6}{7}\right)^0+\left(\frac{1}{2}\right)^2\div2=3-1+\frac{1}{4}\times\frac{1}{2}=2+\frac{1}{8}=\frac{17}{8}\)
Bài 3
a) \(9\times3^3\times\frac{1}{81}\times3^2=3^2\times3^3\times\frac{1}{3^4}\times3^2=3^3\)
b) \(4\times2^5\div\left(2^3\times\frac{1}{16}\right)=2^2\times2^5\div\left(2^3\times\frac{1}{2^4}\right)=2^7\div\frac{1}{2}=2^6\)
c) \(3^2\times2^5\times\left(\frac{2}{3}\right)^2=3^2\times2^5\times\frac{2^2}{3^2}=3^2\times\frac{2^7}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\times\frac{1}{3}\times9^2=\left(\frac{1}{3}\right)^3\times3^4=\frac{1}{3^3}\times3^4=3^1\)
a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)
\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
a, \(\left(2x-1\right)=-8\)
\(2x=-8+1\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
a) (2x - 1) = -8
⇒ 2x = -8 + 1
⇒ 2x = -7
b) (3x - 2)\(^2\) = \(\frac{1}{49}\)
Ta có: \(\frac{1}{49}\) = \(\frac{1}{7}\). \(\frac{1}{7}\) hoặc \(\frac{1}{49}\) = \(\frac{-1}{7}\). \(\frac{-1}{7}\)
TH1: 3x - 2 = \(\frac{1}{7}\) TH2: 3x - 2 = \(\frac{-1}{7}\)
⇒ 3x = \(\frac{1}{7}\)+2 ⇒ 3x = \(\frac{-1}{7}\)+2
⇒ 3x = \(\frac{15}{7}\) ⇒ 3x = \(\frac{13}{7}\)
⇒ x = \(\frac{5}{7}\) ⇒ x = \(\frac{13}{21}\)
Vậy: x = \(\frac{5}{7}\) hoặc x = \(\frac{13}{21}\)
a, \(\frac{x+1}{5}=\frac{3}{7}\Rightarrow7\left(x+1\right)=15\Rightarrow7x+7=15\Rightarrow7x=8\Rightarrow x=\frac{8}{7}\)
b, \(\frac{x-2}{3}=\frac{3}{8}\Rightarrow8\left(x-2\right)=9\Rightarrow8x-16=9\Rightarrow8x=25\Rightarrow x=\frac{25}{8}\)
c, \(\frac{-x-1}{2}=\frac{-3}{5}\Rightarrow5\left(-x-1\right)=-6\Rightarrow-5x-5=-6\Rightarrow-5x=-1\Rightarrow x=\frac{1}{5}\)
d, \(\frac{4}{5-x}=\frac{1}{3}\Rightarrow5-x=12\Rightarrow x=-7\)
e, \(2x\left(x-\frac{1}{7}\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
\(\left(\frac{-7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}=\frac{-7}{4}\cdot\frac{8}{5}\cdot\frac{11}{16}=\frac{-7.11}{4.5.2}=\frac{-77}{40}\)
#Nguồn: Băng
a, \(0,5.8+2^3.\frac{1}{4}-2019^0\)
\(=\frac{1}{2}.8+8.\frac{1}{4}-1\)
\(=4+2-1\)
\(=5\)
b, \(9\left(-\frac{1}{3}\right)^2+\frac{1}{3}-2\left(-\frac{1}{2}\right)^2-\frac{1}{2}\left(\frac{1}{2}\right)^0\)
\(=9.\frac{1}{9}+\frac{1}{3}-2.\frac{1}{4}-\frac{1}{2}.1\)
\(=1+\frac{1}{3}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{1}{3}\)
c, \(\sqrt{25}+\frac{\sqrt{16}}{49}-\sqrt{\left(-3\right)^2}\)
\(=5+\frac{4}{49}-3\)
\(=\frac{42}{19}\)