7x6^10x2^20*3^6-2^19x6^15
Phần
9x6^19x2^9-4x3^17x2^26
Chu y : x co nghia la dau ''nhân''
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a)3/4+1/1/4*2/2/3-(-1/2)^2:6/5
=3/4+5/4*8/3-1/4:6/5
=3/4+10/3-5/24=18/24+80/24-5/24=93/24=31/8
b)(x-1)^5=32=2^5
=>x-1=2
x=2+1
x=3
\(\frac{1}{4}+\frac{3}{8}+\frac{5}{16}=\frac{5}{8}+\frac{5}{16}=\frac{15}{16}\)
\(\frac{3}{5}-\frac{1}{3}-\frac{1}{6}=\frac{4}{15}-\frac{1}{6}=\frac{1}{10}\)
\(\frac{4}{7}\cdot\frac{5}{8}\cdot\frac{7}{12}=\frac{4.5.7}{7.8.12}=\frac{140}{672}=\frac{5}{24}\)
\(\frac{25}{28}:\frac{15}{14}\cdot\frac{6}{7}=\frac{5}{6}\cdot\frac{6}{7}=\frac{5}{7}\)
\(\frac{1}{4}+\frac{3}{8}+\frac{5}{16}=\frac{15}{16}\)
\(\frac{3}{5}-\frac{1}{3}-\frac{1}{6}=\frac{1}{10}\)
\(\frac{4}{7}\times\frac{5}{8}\times\frac{7}{12}=\frac{5}{24}\)
\(\frac{25}{28}:\frac{15}{14}\times\frac{6}{7}=\frac{5}{7}\)
\(\frac{4}{\frac{2}{5}}:\left(-\frac{33}{10}\right)+x=-\frac{1}{\frac{5}{6}}\)
\(10:\left(-\frac{33}{10}\right)+x=-\frac{6}{5}\)
\(-\frac{100}{33}+x=-\frac{6}{5}\)
\(x=\frac{302}{165}\)
Bài 1:
a: \(\frac{5\times7}{15}=\frac{35}{15}=\frac73\)
b: \(8:\frac{16}{23}=8\times\frac{23}{16}=\frac{8}{16}\times23=23\times\frac12=\frac{23}{2}\)
c: \(\frac{8}{15}:5=\frac{8}{15\times5}=\frac{8}{75}\)
d: \(\frac{15}{19}\times3=\frac{15\times3}{19}=\frac{45}{19}\)
Bài 2:
a: \(3\times\frac49+\frac23-\frac{7}{18}\)
\(=\frac43+\frac23-\frac{7}{18}\)
\(=\frac63-\frac{7}{18}=2-\frac{7}{18}=\frac{36}{18}-\frac{7}{18}=\frac{29}{18}\)
b: \(\frac{7}{10}\times\frac23-\frac14=\frac{14}{30}-\frac14=\frac{28}{60}-\frac{15}{60}=\frac{28-15}{60}=\frac{13}{60}\)
c: \(\frac98-\frac78:\frac56=\frac98-\frac78\times\frac65\)
\(=\frac98-\frac{42}{40}=\frac{45}{40}-\frac{42}{40}=\frac{3}{40}\)
d: \(\frac23+\frac12:\frac32-\frac{11}{24}\)
\(=\frac23+\frac12\times\frac23-\frac{11}{24}\)
\(=\frac23+\frac13-\frac{11}{24}=1-\frac{11}{24}=\frac{13}{24}\)
Bài 3:
Độ dài đáy tương ứng với chiều cao là: \(\frac35\times2=\frac65\left(m\right)\)
Chu vi hình bình hành là:
\(\left(\frac65+\frac45\right)\times2=\frac{10}{5}\times2=2\times2=4\left(m\right)\)
Diện tích bình hành là;
\(\frac65\times\frac35=\frac{18}{25}\left(m^2\right)\)
\(a,x^4-4x^3-19x^2+106x-120=0\\ \Rightarrow\left(x-4\right)\left(x^3-19x+30\right)=0\Rightarrow\left(x-4\right)\left(x+5\right)\left(x^2-5x+6\right)=0\\ \Rightarrow\left(x-4\right)\left(x+5\right)\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-5\\x=2\\x=3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-5;2;3;4\right\}\)
\(b,4x^4+12x^3+5x^2-6x-15=0\\ \Rightarrow\left(x-1\right)\left(4x^3+16x^2+21x+15\right)=0\\ \Rightarrow\left(x-1\right)\left[\left(4x^3+10x^2\right)+\left(6x^2+15x\right)+\left(6x+15\right)\right]=0\\ \Rightarrow\left(x-1\right)\left[2x^2\left(2x+5\right)+3x\left(2x+5\right)+3\left(2x+5\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(2x+5\right)\left(2x^2+3x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\\2x^2+3x+3=0\left(vô.lí\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;-\dfrac{5}{2}\right\}\)