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Câu 1:
\(a,=43\cdot\left(27+93\right)+3111+3363=43\cdot120+6474=11634\\ b,=11^2+2^{15}\cdot2^3:2^{17}=121+2=123\\ c,=11^2+7^2-9=121+49-9=151\)
Câu 2:
\(a,\Rightarrow x-\dfrac{3}{2}=5^2=25\\ \Rightarrow x=25+\dfrac{3}{2}=\dfrac{53}{2}\\ b,\Rightarrow7x=30-2=28\\ \Rightarrow x=4\)
1,
a,\(2020-\left(249+2020\right)+\left(249-573\right)\)
\(=2020-249-2020+249-573\)
\(=-573\)
b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)
\(=257+1-18-257\)
\(=1-18=-17\)
\(c,25.\left(85-47\right)-85.\left(47+25\right)\)
\(=25.85-25.47-47.85+85.25\)
\(=85.\left(25-47+25\right)-25.47\)
\(=85.3-25.47\)
\(=-920\)
2,
\(a,15-5.\left(x+2\right)=-30\)
\(=>5.\left(x+2\right)=15+30=45\)
\(=>x+2=\frac{45}{5}=9\)
\(=>x=7\)
\(b,\left(x+2\right)^2+5=105\)
\(=>\left(x+2\right)^2=100\)
\(=>\left(x+2\right)^2=10^2\)
\(=>x+2=10\)
\(=>x=8\)
\(c,\left|2x-5\right|-\left(-6\right)=11\)
\(=>\left|2x-5\right|=11-6=5\)
\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)
\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)
a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)
=>7-x+5=15-x+x-12
=>12-x=3
hay x=9
b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)
\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)
=>x-38-x=13-40-x
=>-27-x=-38
=>x+27=38
hay x=11
e: \(x^2+3x+9⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)
hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)
; b) 
;
; d) 
;
Bài 1:
a) Ta có: 49+(11-25)
=49+11-25
=24+11
=35
b) Ta có: \(-8+5\cdot\left(-9\right)\)
\(=-8-45\)
=-53
Bài 2:
a) Ta có: \(x-2=-6+17\)
\(\Leftrightarrow x-2=11\)
hay x=13
Vậy: x=13
b) Ta có: \(x+9=2-17\)
\(\Leftrightarrow x+9=-15\)
hay x=-24
Vậy: x=-24
`Answer:`
a. \(-\frac{17}{30}-\frac{11}{-15}+-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{11}{15}-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{22}{30}-\frac{7}{12}\)
\(=\frac{1}{6}-\frac{7}{12}\)
\(=-\frac{5}{12}\)
b. \(-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=\frac{5}{9}.\left(-1\right)+\frac{5}{9}.\left(-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\left(-1-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\frac{-17}{5}\)
\(=-\frac{17}{9}\)
c. \(-\frac{7}{25}.\frac{11}{13}+-\frac{7}{25}-\frac{2}{13}-\frac{18}{25}\)
\(=\frac{77}{325}-\left(\frac{7}{25}+\frac{18}{25}\right)-\frac{2}{13}\)
\(=\frac{77}{325}-1-\frac{2}{13}\)
\(=-\frac{452}{325}\)