a)\(568-\left\{5.\text{[}143-\left(4-1\right)^2\text{]}+10\right\}:10\)

\(=568-\left\{5.\left[143-3^2\right]+10\right\}:10\)

\(=568-\left\{5.134+10\right\}\text{ }\)

\(=568-\left(670+10\right)\)

\(=568-680\)

\(=-112\)

b)\(10^2-\left[60:\left(5^6:5^4-3\times5\right)\right]\)

\(=100-\left[60-\left(6^2-15\right)\right]\)

\(=100-\left(60-\left(36-15\right)\right)\)

\(=100-60-36+15\)

\(=19\)

\(a)\)\(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)

\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)

\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)

\(=\)\(568-\left\{5.134+10\right\}\div10\)

\(=\)\(568-\left\{670+10\right\}\div10\)

\(=\)\(568-680\div10\)

\(=\)\(568-68\)

\(=\)\(500\)

\(b)\)\(10^2-\left[60\div\left(5^6\div5^4-3\times5\right)\right]\)

\(=\)\(10^2-\left[60\div\left(5^2-15\right)\right]\)

\(=\)\(10^2-\left[60\div\left(25-15\right)\right]\)

\(=\)\(10^2-\left[60\div10\right]\)

\(=\)\(100-6\)

\(=\)\(94\)

Bn chỉ cần bít tính và bít nhân chia trước cộng trừ sau!

Bn tỉnh ngoài tập nhap nhé!

Tìm x thì bn chỉ thuộc công thức nhé!

Và bít thư lại tìm x,các phép tính.

k cho mìh nhé

2*3+6789=13395                 12345*5=61725                    x*456=912
987*3+78=3039                   4567*3=13701                            x=912:456
56+43*8=400                       13200*3=39600                          x=2
987*4+45=4002                    567*4=2268
[45*3]+98=233                     7895+2349=10244                568*x=1136
[56+45]*5=505                     4567-2399=2168                         x=1136:568
245:5*7=343                                                                          x=2
456*3=1368
789*9=7091
67*9=603
8*9=72

a) (x – 45).27 = 0 ó x – 45 = 0 ó x = 45

b) 45.(2x – 4).13 = 0 ó 2x – 4 = 0 ó 2x – 4 = 0 ó 2x = 4 ó x = 2

c) (x – 3).(x – 5) = 0

Vậy tập nghiệm của phương trình là S = {3;5}

a x=45+0=45

a: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)

=>\(5-\sqrt{2x-1}=0\)

=>\(\sqrt{2x-1}=5\)

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

=>\(\sqrt{\left(x-5\right)^2}=2\)

=>\(\left|x-5\right|=2\)

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

=>\(\sqrt{\left(x-7\right)^2}=5\)

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)

\(6^2-\left(x+3\right)=45\)

\(36-\left(x+3\right)=45\)

\(x+3=35-45\)

\(x+3=-10\)

\(x=-13\)

lớp 6 mà bạn

\(6^2-\left(x+3\right)=45\)

\(\Rightarrow36-\left(x+3\right)=45\)

\(\Rightarrow x+3=36-45\)

\(\Rightarrow x+3=-9\)

\(\Rightarrow x=-9-3=-12\)

\(125-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-15x+5=5^5.5^3\)

\(\Rightarrow5\left(5^2-3x+1\right)=5^5.5^3\)

\(\Rightarrow5^2-3x+1=5^5.5^3:5\)

\(\Rightarrow25-3x+1=5^7\)

Từ đây làm nốt nhé

\(4^x+1+4^0=65\)

\(\Rightarrow4^{x+1}+1=65\)

\(\Rightarrow4^{x+1}=64\)

\(\Rightarrow4^{x+1}=4^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

\(70-5\left(x-3\right)=5.3^2\)

\(\Rightarrow70-5\left(x-3\right)=5.9=45\)

\(\Rightarrow5\left(x-3\right)=70-45=25\)

\(\Rightarrow x-3=25:5=5\)

\(\Rightarrow x=8\)

Chúc em học tốt hơn nhé!!

496

    \(568-\left\{5.\left[143-\left(4-1\right)^2+10\right]\right\}:10\)

\(=568-\left\{5.\left[143-3^2+10\right]\right\}:10\)

\(=568-\left\{5.\left[143-9+10\right]\right\}:10\)

\(=568-\left\{5.144\right\}:10=568-720:10=568-72=496\)

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