a.74×42+74×58-400
b.75-(3×5^2-4×2^3)
c)x-15=-135
help me😭😭😭😭
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a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)
\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)
\(\Leftrightarrow19x-21=-27\)
=>19x=-6
hay x=-6/19
b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2=\dfrac{3}{2}\)
\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)
\(\Leftrightarrow3x^2+22x+16=0\)
\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)
\(\Leftrightarrow10x^2-19x=0\)
=>x(10x-19)=0
=>x=0 hoặc x=19/10
a) 3571 - ( 1265 + x ) = 728 : 14
3571 - ( 1265 + x ) = 52
1265 + x = 3571 - 52
1265 + x = 3519
x = 3519 - 1265
x = 2254
b) ( 15 + 4x ) : 29 + 32 = 47
( 15 + 4x ) : 29 = 47 - 32
( 15 + 4x ) : 29 = 15
15 + 4x = 15 . 49
15 + 4x = 735
4x = 735 - 15
4x = 720
x = 180
2 câu dễ làm trước, 2 câu còn lại tối đi học về mới làm được..(giờ bận rồi)
a) ĐẶt \(x^2+3x+1=a\)
\(A=a\left(a-4\right)-5=a^2-4a-5=\left(a-5\right)\left(a+1\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
c)\(C=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt ẩn phụ: \(t=x^2+8x+7\) rồi làm tiếp đi..
Để anh làm nốt vậy.
\(B=\left(x^2+2x\right)^2-2x^2-4x-3\)
\(B=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1-4\)
\(B=\left(x^2+2x-1\right)^2-2^2\)
\(B=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
\(B=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
___
\(D=x^2-2xy+y^2-7x+7y+12\)
\(D=\left(x-y\right)^2-7\left(x-y\right)+12\)
\(D=\left(x-y\right)^2-3\left(x-y\right)-4\left(x-y\right)+12\)
\(D=\left(x-y\right)\left(x-y-3\right)-4\left(x-y-3\right)\)
\(D=\left(x-y-3\right)\left(x-y-4\right)\)
a) Ta có: 2|x + 2| \(\ge\)0 \(\forall\)x
=> 2|x + 2| + 15 \(\ge\)15 \(\forall\)x
Hay A \(\ge\)15 \(\forall\)x
Dấu "=" xảy ra <=>x + 2 = 0 <=> x = -2
Vậy Min A = 15 tại x = -2
b) Ta có: 2(x + 5)4 \(\ge\)0 \(\forall\)x
3|x + y + 2| \(\ge\)0 \(\forall\)x;y
=> 20 - 2(x + 5)4 - 3|x + y + 2| \(\le\)20 \(\forall\)x;y
Hay B \(\le\)20 \(\forall\)x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+5=0\\x+y+2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-5\\y=-2-x\end{cases}}\) <=> \(\hept{\begin{cases}x=-5\\y=-2-\left(-5\right)=3\end{cases}}\)
Vậy Max B = 20 tại x = -5 và y = 3
A. 1/7 + 5/14==......=.......;
B. 3/4 - 5/12=..............=......=.......;
C. 4/15 × 5/14=...........=......=.......;
D. 3/4 : 5/12=..............=......=.........
\(A=1+2+2^2+2^3+...+2^{2020}\)
\(2A=2+2^2+2^3+2^4+...+2^{2021}\)
\(2A-A=\left(2+2^2+2^3+2^4+....+2^{2021}\right)-\left(1+2+2^2+2^3+...+2^{2020}\right)\)
\(A=2^{2021}-1\)
a, 15/x - 1/3 = 28/57
15/x = 28/57 + 1/3
15/x = 28/57 + 19/57
15/x = 47/57
x . 47 = 15 . 57
x = 855/47
b, x/2 - 2/5 = 1/10
x/2 = 1/10 + 2/5
x/2 = 1/10 + 4/10
x/2 = 5/10 = 1/2
x/2 = 1/2
=> x=1
a) \(74.42+74.58-400=74.\left(42+58\right)-400\)
\(=74.100-400\)
\(=7400-400\)
\(=7000\)
b) \(75-\left(3.5^2-4.2^3\right)=75-\left(3.25-4.8\right)\)
\(=75-\left(75-32\right)\)
\(=75-75+32\)
\(=32\)
c) \(x-15=-135\)
\(x=-135+15\)
\(x=-120\)