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7 tháng 8 2019

\(a,347\cdot2^2-2^2\cdot\left(216+184\right):8\)

\(< =>1388-4\cdot400:8\)

\(< =>1388-1600:8\)

\(< =>1388-200\)

\(< =>1188\)

\(b,132-\left[116-\left(132-128\right)^2\right]\)

\(< =>132-\left[116-\left(4^2\right)\right]\)

\(< =>132-100\)

\(< =>32\)

\(c,\left[184:\left(96-124:31\right)-2\right]\cdot3651\)

\(< =>\left[184:\left(96-4\right)-2\right]\cdot3651\)

\(< =>\left[184:92-2\right]\cdot3651\)

\(< =>\left[2-2\right]\cdot3651\)

\(< =>0\cdot3651\)

\(< =>0\)

\(e,\left(2+4+6+8+...+2020\right)\cdot\left(36\cdot333-108\cdot111\right)\)

\(< =>\left[\left(2020-2\right):2+1\right]\cdot\left(36\cdot333-108\cdot111\right)\)

\(< =>1010\cdot\left(11988-11988\right)\)

\(< =>1010\cdot0\)

\(< =>0\)

\(g,1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)

\(< =>64+140:70-49\)

\(< =>64+2-49\)

\(< =>66-49\)

\(< =>17\)

\(h,\left(44\cdot52\cdot60\right):\left(11\cdot13\cdot15\right)\)

\(< =>137280:2145\)

\(< =>64\)

\(j,\left(2^{17}+15^4\right)\cdot\left(3^{19}-2^{17}\right)\cdot\left(2^4-4^2\right)\)

\(< =>\left(131072+50625\right)\cdot\left(1162261467-131072\right)\cdot\left(16-16\right)\)

\(< =>181697\cdot1162130395\cdot0\)

\(< =>0\)

mk chuc ban hoc tot nhe :))

3 tháng 4 2018

Căng, sự thật là nó rất căng

Nhg dù sao thì.....

1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)

Xét \(A\left(x\right)=0\)

\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)

\(\Rightarrow-3x^2-12x+15=0\)

\(\Rightarrow-3x^2+3x-15x+15=0\)

\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)

Xét \(B\left(x\right)=0\)

\(\Rightarrow x^3+x^2-4x-4=0\)

\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)

Đó là những j mình biết khocroikhocroi

5 tháng 2 2018

1, \(\left(x-4\right)^2-\left(2x+1\right)^2=\left(x-4-2x-1\right)\left(x-4+2x+1\right)=-3\left(x+5\right)\left(x-1\right).\)

\(\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)(mấy cái này áp dụng hàng đẳng thức lớp 8 mới hok)

2,\(x^3+x^2-4x-4=\left(x-2\right)\left(x^2+3x+2\right)=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)

\(\orbr{\begin{cases}x=\mp2\\\end{cases}}x=-1\)

tương tụ lm tiếp nhe buồn ngủ quá rồi !

c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)

\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)

d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)

e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)

i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)

\(=3^{40}-1\)

c: \(=\dfrac{7}{23}\cdot\dfrac{-24-45}{18}=\dfrac{7}{23}\cdot\dfrac{-69}{18}=\dfrac{7}{18}\cdot\left(-3\right)=-\dfrac{7}{6}\)

d: \(=\dfrac{7}{5}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\)

e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)

i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{9^5}=3^{40}-1\)

14 tháng 7 2016

4) \(2.3^x+3^{x-1}=7.\left(3^2+2.6^2\right)\)

\(\Rightarrow2.3^x+3^{x-1}=567\)

\(\Rightarrow7.3^{x-1}=567\)

\(\Rightarrow3^{x-1}=567\div7\)

\(\Rightarrow3^{x-1}=81\)

\(\Rightarrow3^{x-1}=3^4\)

\(\Rightarrow x-1=4\)

\(\Rightarrow x=4+1\)

\(\Rightarrow x=5\)

Vậy \(x=5\)