(1,5 điểm) Tìm x :
a) x × 7 + 85 = 106
b) x 9 = 819
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a) x – 258 = 347
x = 347 + 258
x = 605
b) x × 9 = 819
x = 819 : 9
x = 91
Bài 1:
a, 58.32 + 58.68 - 800
= 58.(32 + 68) - 800
= 58.100 - 800
= 5800 - 800
= 5000
b, 12020 + 280 : [55 - (7 - 4)3]
= 12020 + 280 : [ 55 - 33 ]
= 12020 + 280 : [ 28]
= 12020 + 10
= 12030
c, (96 - 19 - 45) - (55 + 96 - 119)
= 96 - 19 - 45 - 55 - 96 + 119
= (96 - 96) + (119 - 19) - (45 + 55)
= 0 + 100 - 100
= 0
2:
a: x+201=351
=>x=351-201
=>x=150
b: \(8\cdot5^2-27:25\)
\(=8\cdot25-\dfrac{27}{25}\)
\(=200-1,08=198,92\)
d: \(2023-23:\left[9+2\left(2^3-0,21\right)\right]\)
\(=2023-23:\left[9+16-0,42\right]\)
\(=2023-\dfrac{23}{25-0,42}\)
\(=2023-\dfrac{1150}{1229}=\dfrac{2485117}{1229}\)
b: 2(x-21)=84
=>x-21=84/2=42
=>x=42+21=63
c: 135-4(81-x)=55
=>4(81-x)=135-55=80
=>81-x=20
=>x=61
( 18 - 9x ) x 7 = 819 - 756
( 18 - 9x ) x 7 = 63
18 - 9x = 63 : 7
18 - 9x = 9
9x = 18 - 9
9x = 9
x = 9 : 9
x = 1
( 18 - 9\(x\) ) x 7 = 819 - 756
=>7x18 - 7x 9\(x\)= 63
=>126 - 63\(x\)= 63
=>63\(x\)= 126 - 63
=>63\(x\)= 63
=>\(x\)= 36 : 36 = 1
OK
a: \(A=\left|3x-9\right|+1.5\ge1.5\forall x\)
Dấu '=' xảy ra khi x=3
b: \(B=\left|x-7\right|-14\ge-14\forall x\)
Dấu '=' xảy ra khi x=7
a,x/9=5/3 = x=15
b,17/x=85/105 = x = 21
c,6/8=15/x = x = 20
d,x/8=2/3 = x = 6
a: Thay x=49 vào A, ta được:
\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)
b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)
P>1 khi P-1>0
=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)
=>\(\sqrt{x}-2>0\)
=>\(\sqrt{x}>2\)
=>x>4
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)
\(\frac{7.x}{8}-5.\left(x-9\right)=20.x+\frac{1,5}{6}\)
\(\Leftrightarrow7.x-5.\left(x-9\right)=20x.8+\frac{1,5}{6}.8\)
\(\Leftrightarrow7.x-40\left(x-9\right)=160x+2\)
\(\Leftrightarrow7x-40+360=160x+2\)
\(\Leftrightarrow-33x+360=160x+2\)
\(\Leftrightarrow-33x=160x+x-360\)
\(\Leftrightarrow-33x=160x-358\)
\(\Leftrightarrow-33x-160x=160x-358\)
\(\Leftrightarrow-193x=-358\)
\(\Rightarrow x=\frac{358}{193}\)
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
Mỗi câu đúng được 0,75 điểm
a) x × 7 + 85 = 106
x × 7 = 106 – 85
x × 7 = 21
x = 21 : 7
x = 3
b) x : 9 + 72 = 137
x : 9 = 137 - 72
x : 9 = 65
x = 65 x 9
x = 585
con cai nit